# Preliminary Classification of Sub-case (3)

• Xu Bangteng
• Guiyun Chen
• Effi Cohen
• Arisha Haj Ihia Hussam
• Mikhail Muzychuk
Chapter
Part of the Algebra and Applications book series (AA, volume 16)

## Abstract

In this chapter the case (3) is studied in more details. It is shown that if the element b 10 appearing in the product b 3 b 8 is non-real, then b 3 b 10 contains two constituents one of which is a non-real element of degree 15.

## 4.1 Introduction

In Chap. 4 we shall freely use the definitions and notation used in Chap. . The Main Theorem 2 of Chap.  left two unsolved problems for the complete classification of the Normalized Integral Table Algebra (NITA) (A,B) generated by a faithful non-real element of degree 3 with L 1(B)=1 and L 2(B)=∅. In this chapter we solve the second problem where (b 3 b 10,b 3 b 10)≠2, and prove that the NITA that satisfies case (2) of the Main Theorem 1 of Chap.  where (b 3 b 10,b 3 b 10)≠2 does not exist. Consequently, we can state the Main Theorem 3 of this chapter as follows.

### Main Theorem 3

Let (A,B) be a NITA generated by a non-real element b 3B of degree 3 and without non-identity basis element of degree 1 and 2. Then $$b_{3} \bar{b}_{3} = 1 + b_{8}$$, b 8B, and one of the following holds:
1. (1)

There exists a real element b 6B such that $$b_{3}^{2} =\bar{b}_{3} + b_{6}$$ and (A,B)≅ x (CH(PSL(2,7)),Irr(PSL(2,7))).

2. (2)

There exist b 6,b 10,b 15B, where b 6 is non-real, such that $$b_{3}^{2} = \bar{b}_{3} + b_{6}$$, $$\bar{b}_{3} b_{6} = b_{3} + b_{15}$$, b 3 b 6=b 8+b 10, and (b 3 b 8,b 3 b 8)=3. Moreover, if b 10 is real then (A,B)≅ x (CH(3⋅A 6),Irr(3⋅A 6)) of dimension 17, and if b 10 is non-real then (b 3 b 10,b 3 b 10)=2 and b 15 is a non-real element.

3. (3)

There exist c 3,b 6B, c 3b 3 or $$\bar{b}_{3}$$, such that $$b_{3}^{2} = c_{3} + b_{6}$$ and either (b 3 b 8,b 3 b 8)=3 or 4. If (b 3 b 8,b 3 b 8)=3 and c 3 is non-real, then (A,B)≅ x (A(3⋅A 6⋅2),B 32) of dimension 32. (See Theorem 2.9 of Chap for the definition of this specific NITA.) If (b 3 b 8,b 3 b 8)=3 and c 3 is real, then (A,B)≅ x (A(7⋅5⋅10),B 22) of dimension 22. (See Theorem 2.10 of Chap for the definition of this specific NITA.)

In the above Main Theorem 3 we still have 2 open problems in the cases (2) and (3). In case (2) we must classify the NITA such that b 10 is non-real and (b 3 b 10,b 3 b 10)=2, and in case (3) we must classify the NITA such that (b 3 b 8,b 3 b 8)=4. In Chap. , the open case (2) will be solved.

Let us emphasize that the NITA’s of dimension 7 and 17 in the cases (1) and (2) of the Main Theorem 2 are strictly isomorphic to the NITA’s induced from finite groups G via the basis of the irreducible characters of G. However, the NITA’s of dimensions 22 and 32 are not induced from finite groups as described in Chap. .

The Main Theorem 3 follows from the Main Theorem 2 in Chap.  and the next theorem.

### Theorem 4.1

Let (A,B) be a NITA generated by a non-real element b 3B of degree 3 and without non-identity basis element of degree 1 or 2. Then $$b_{3}\bar{b}_{3}=1+b_{8}$$, b 8B. Assume that
$$b_3^2=\bar{b}_3+b_6, \quad \bar{b}_3b_6=b_3+b_{15},\quad b_3b_6=b_8+b_{10},\quad \hbox {and} \quad b_3b_8=b_3+\bar{b}_6+b_{15}$$
where b 6, b 10 are non-real elements in B and b 15B. Then b 15 is a non-real element and (b 3 b 10,b 3 b 10)=2.

The rest of this chapter is devoted to proving the above theorem.

## 4.2 Preliminary Results

For the rest of this chapter, we shall always assume that (A,B) is a NITA generated by a non-real element b 3B of degree 3 and without non-identity basis element of degree 1 and 2 such that
$$b_3^2= \bar b_3+ b_6 ,$$
(4.1)
$$\begin{array}{*{20}l} {b_3 \bar b_3= 1 + b_8 ,} & {b_8\in B,}\\ \end{array}$$
(4.2)
$$\begin{array}{*{20}l} {\bar b_3 b_6= b_3+ b_{15} ,} & {b_{15}\in B,}\\ \end{array}$$
(4.3)
$$\begin{array}{*{20}l} {b_3 b_6= b_8+ b_{10} ,} & {b_{10}\in B,}\\ \end{array}$$
(4.4)
and
$$b_3b_8=b_3+\bar{b}_6+b_{15}.$$
(4.5)
By (4.4), (4.5) and $$\bar{b}_{3}(b_{3}b_{6})=(b_{3}\bar{b}_{3})b_{6}$$, we obtain that
$$b_6b_8=\bar{b}_3+\bar{b}_{15}+\bar{b}_3b_{10}.$$
(4.6)
By (4.1), (4.3), (4.4), (4.5) and $$b_{3}(b_{3}b_{6})=b_{3}^{2}b_{6}$$ we obtain that
$$b_6^2=\bar{b}_6+b_3b_{10}.$$
(4.7)
By (4.2) and (4.3), we obtain that
$$(\bar{b}_3b_6,b_3b_{10})=(b_3,b_3b_{10})+(b_{15},b_3b_{10})=(b_{15},b_3b_{10});$$
on the other hand, (4.1) and (4.4) imply that
$$(\bar{b}_3b_6,b_3b_{10})=(b_3^2,b_6\bar{b}_{10})=(\bar{b}_3,b_6\bar{b}_{10})+(b_6,b_6\bar{b}_{10})=1+(b_6,b_6\bar{b}_{10}).$$
Hence $$1+(b_{6},b_{6}\bar{b}_{10})=(b_{15},b_{3}b_{10})$$ which implies that
$$b_3b_{10}=b_{15}+R_{15} \quad\hbox{where } R_{15}\in \mathbb{N}B.$$
(4.8)
The degrees appearing in R 15 must all be ≥5.
By (4.1), (4.3), (4.5) and $$b_{3}(b_{3}b_{8})=b_{3}^{2}b_{8}$$, we obtain that
$$b_6b_8=\bar{b}_3+b_3b_{15}.$$
(4.9)
Now (4.6) implies that
$$b_3b_{15}=\bar{b}_{15}+\bar{b}_3b_{10}.$$
(4.10)
By (4.1), (4.2), (4.3), (4.4) and $$(b_{3}\bar{b}_{3})(b_{3}\bar{b}_{3})=b_{3}^{2}\bar{b}_{3}^{2}$$, we obtain that
$$b_8^2=b_8+b_{10}+\bar{b}_{10}+b_6\bar{b}_6.$$
(4.11)
By (4.1), (4.2), (4.3), (4.4) and $$\bar{b}_{6}b_{3}^{2}=(b_{3}\bar{b}_{6})b_{3}$$, we obtain that
$$\bar{b}_{10}+b_6\bar{b}_6=1+b_3\bar{b}_{15}.$$
(4.12)
By (4.1), (4.8), (4.10) and $$b_{3}(b_{3}b_{10})=b_{3}^{2}b_{10}$$, we obtain that
$$b_6b_{10}=\bar{b}_{15}+b_3R_{15}.$$
(4.13)
The proof that b 15 is non-real is simple.
Assume henceforth that b 15 is a real element and we shall derive a contradiction. By (4.3) we have that (b 6,b 3 b 15)=1 and together with (4.12), we obtain that $$(\bar{b}_{6},b_{3}b_{15})=1$$ which implies that (b 15,b 3 b 6)=1, and we have a contradiction to (4.4). Therefore
$$\hbox{b_{15}\quad is a non-real element}.$$
(4.14)

The remainder of this chapter will consist of the proof that (b 3 b 10,b 3 b 10)=2.

By (4.4) we obtain that $$(b_{6},\bar{b}_{3}b_{10})=1$$ which implies that (b 3 b 10,b 3 b 10)≥2. By (4.4) we obtain that $$(\bar{b}_{6},b_{3}b_{10})=0$$. Now by (4.3), (4.4) and (4.7) we obtain that $$(b_{3}b_{10},b_{3}b_{10})=(b_{6}^{2},b_{3}b_{10})=(\bar{b}_{3}b_{6},\bar{b}_{6}b_{10})=(b_{3},\bar{b}_{6}b_{10})+(b_{15},\bar{b}_{6}b_{10})=1+(b_{15},\bar{b}_{6}b_{10})$$. Since $$(b_{3},\bar{b}_{6}b_{10})=1$$, we have that $$(b_{15},\bar{b}_{6}b_{10})\leq 3$$ which implies that (b 3 b 10,b 3 b 10)≤4. Therefore
$$2\leq (b_3b_{10},b_3b_{10})\leq 4.$$
(4.15)

Assume henceforth that b 15=R 15 and we shall derive a contradiction. Then by (4.8) we obtain that b 3 b 10=2b 15 which implies that $$(\bar{b}_{10},b_{3}\bar{b}_{15})=2$$, and by (4.12) we obtain that $$(\bar{b}_{10},b_{6}\bar{b}_{6})=1$$ which implies that $$(b_{10},b_{6}\bar{b}_{6})=1$$. Now (4.12) implies that $$(\bar{b}_{15},\bar{b}_{3}b_{10})=(b_{10},b_{3}\bar{b}_{15})=1$$. By (4.4) we obtain that $$(b_{6},\bar{b}_{3}b_{10})=1$$ and since (b 3 b 10,b 3 b 10)=4, we obtain that $$\bar{b}_{3}b_{10}=b_{6}+\bar{b}_{15}+\alpha+\beta$$ where α,βB, |α|+|β|=9 and |α|,|β|≥5, which is a contradiction to our assumption that b 15=R 15.

Now by (4.8) and (4.15), we have five cases (see Table 4.1).
Table 4.1

Splitting to five cases

 Case 1 R 15=x 5+x 10, x 5,x 10∈B Case 2 R 15=x 6+x 9, x 6,x 9∈B Case 3 R 15=x 7+x 8, x 7,x 8∈B Case 4 R 15=x 5+y 5+z 5, x 5,y 5,z 5∈B Case 5 R 15∈B, b 15≠R 15

In the rest of this chapter, we derive a contradiction for Cases 1–4 in the above table. In Case 5, by (4.8) we then have that (b 3 b 10,b 3 b 10)=(b 15,b 15)+(R 15,R 15)=2. Then we will have proven Theorem 1.1.

## 4.3 Case R15=x5+x10

In this section we assume that R 15=x 5+x 10. Then by (4.7), (4.8) and (4.13), we have that
$$b_3 b_{10}= b_{15}+ x_5+ x_{10} ,$$
(4.16)
$$b_6^2= \bar b_6+ b_{15}+ x_5+ x_{10}$$
(4.17)
and
$$b_6b_{10}=\bar{b}_{15}+b_3x_5+b_3x_{10}.$$
(4.18)
Now by (4.16) we obtain that $$(b_{10},\bar{b}_{3}x_{5})=1$$ which implies that
$$\bar{b}_3x_5=b_{10}+y_5,\quad \hbox {where }y_5\in B.$$
(4.19)
Now (4.1) and $$\bar{b}_{3}^{2}x_{5}=\bar{b}_{3}(\bar{b}_{3}x_{5})$$ imply that
$$b_3x_5+\bar{b}_6x_5=\bar{b}_3b_{10}+\bar{b}_3y_5.$$
(4.20)
By (4.2), (4.16), (4.19) and $$b_{3}(\bar{b}_{3}x_{5})=(b_{3}\bar{b}_{3})x_{5}$$, we obtain that
$$b_8x_5=x_{10}+b_{15}+b_3y_5.$$
(4.21)
Now by (4.4), (4.19) and $$(\bar{b}_{3} \bar{b}_{6})x_{5}=\bar{b}_{6}(\bar{b}_{3}x_{5})$$, we obtain that
$$\bar{b}_6b_{10}+\bar{b}_6y_5=x_{10}+b_{15}+b_3y_5+\bar{b}_{10}x_5.$$
(4.22)
By (4.1) and (4.19) we obtain that $$(\bar{b}_{3}y_{5},b_{3}x_{5})=(b_{3}^{2},y_{5}\bar{x}_{5})=(\bar{b}_{3},y_{5}\bar{x}_{5})+(b_{6},y_{5}\bar{x}_{5})=1+(b_{6},\bar{x}_{5}y_{5})$$. Hence
$$(\bar{b}_3y_5,b_3x_5)=1+(b_6,\bar{x}_5y_5).$$
(4.23)
By (4.2) and (4.19) we obtain that $$2=(b_{3}x_{5},b_{3}x_{5})=(b_{3}\bar{b}_{3},x_{5}\bar{x}_{5})=1+(b_{8},x_{5}\bar{x}_{5})$$. Hence
$$(b_8,x_5\bar{x}_5)=1$$
(4.24)
and together with $$b_{10}\neq \bar{b}_{10}$$, we obtain that
$$(b_{10},x_5\bar{x}_5)=0.$$
(4.25)
Now (4.4) implies that $$(\bar{b}_{3}\bar{x}_{5},b_{6}\bar{x}_{5})=(b_{3}b_{6},x_{5}\bar{x}_{5})=(b_{8},x_{5}\bar{x}_{5})+(b_{10},x_{5}\bar{x}_{5})=1$$. Hence
$$(\bar{b}_3\bar{x}_5,b_6\bar{x}_5)=1.$$
(4.26)
By (4.3) and (4.19) we obtain that $$(\bar{b}_{3}y_{5},\bar{b}_{6}x_{5})=(b_{3}\bar{b}_{6},y_{5}\bar{x}_{5})=(\bar{b}_{3},y_{5}\bar{x}_{5})+(\bar{b}_{15},y_{5}\bar{x}_{5})=1+(\bar{b}_{15},y_{5}\bar{x}_{5})$$. Therefore
$$(\bar{b}_3y_5,\bar{b}_6x_5)=1+(\bar{b}_{15},y_5\bar{x}_5)$$
(4.27)
which implies that
$$(\bar{b}_3y_5,\bar{b}_6x_5)\leq 2.$$
(4.28)
By (4.2) we obtain that $$(b_{3}y_{5},b_{3}y_{5})=(b_{3}\bar{b}_{3},y_{5}\bar{y}_{5})=1+(b_{8},y_{5}\bar{y}_{5})$$ which implies that
$$(b_3y_5,b_3y_5)\leq 4.$$
(4.29)
Assume henceforth that $$(\bar{b}_{3}y_{5},\bar{b}_{3}y_{5})=4$$ and we shall derive a contradiction. Then
$$\bar{b}_3y_5=x+y+z+v\quad\hbox{where } x,y,z,v\in B.$$
Now, since (b 3 x 5,b 3 x 5)=2, by (4.23) we have that $$(b_{3}x_{5},\bar{b}_{3}y_{5})=1$$. Hence
$$b_3x_5=x+w \quad\hbox{where }w\in B,$$
and together with (4.20), we get that
$$x+w+\bar{b}_6x_5=\bar{b}_3b_{10}+x+y+z+v.$$
Thus $$(\bar{b}_{6}x_{5},\bar{b}_{3}y_{5})\geq 3$$, and we have a contradiction to (4.28).
Assume henceforth that $$(\bar{b}_{3}y_{5},\bar{b}_{3}y_{5})=3$$ and we shall derive a contradiction. Then
$$\bar{b}_3y_5=x+y+z\quad\hbox{where } x,y,z\in B$$
and
$$b_3x_5=x+w\quad\hbox{where } w\in B.$$
By (4.20) we have that
$$x+w+\bar{b}_6x_5=\bar{b}_3b_{10}+x+y+z.$$
Now (4.28) implies that $$(x,\bar{b}_{6}x_{5})=0$$ and by (4.26), we obtain that $$(b_{3}x_{5},\bar{b}_{6}x_{5})=1$$ which implies that $$(w,\bar{b}_{6}x_{5})=1$$ and $$(w,\bar{b}_{3}b_{10})=2$$. Since (b 3 b 10,b 3 b 10)=3 we have a contradiction.
Assume henceforth that $$(\bar{b}_{3}y_{5},\bar{b}_{3}y_{5})=1$$ and we shall derive a contradiction. Then $$\bar{b}_{3}y_{5}=c_{15}$$ and by (4.23) we obtain that b 3 x 5=c 15, which is a contradiction to (4.19). Now (4.29) implies that
$$(b_3y_5,b_3y_5)=2.$$
(4.30)
Assume henceforth that $$(\bar{b}_{15},y_{5}\bar{x}_{5})=1$$ and we shall derive a contradiction. Then (4.27) implies that $$(\bar{b}_{3}y_{5},\bar{b}_{6}x_{5})=2$$. Now (4.17) and (4.20) imply that
$$\bar{b}_6x_5=b_6+2x_{12}\quad\hbox{where } x_{12}\in B$$
and
$$\bar{b}_3y_5=x_{12}+c_3\hbox { where } c_3\in B.$$
Now by (4.4) we obtain that $$(b_{6},\bar{b}_{3}\bar{x}_{5})=0$$, which is a contradiction to (4.26). Hence
$$(\bar{b}_{15},y_5\bar{x}_5)=0.$$
(4.31)
By (4.21) and (4.30) we obtain that (b 8 x 5,b 3 y 5)≥2. In addition, (4.5), (4.19) and (4.31) imply that $$(b_{8}x_{5},b_{3}y_{5})=(\bar{b}_{3}b_{8},y_{5}\bar{x}_{5})=(\bar{b}_{3},y_{5}\bar{x}_{5})+(b_{6},y_{5}\bar{x}_{5})+(\bar{b}_{15},y_{5}\bar{x}_{5})=1+(b_{6},y_{5}\bar{x}_{5})$$. Hence
$$(b_6,y_5\bar{x}_5)\geq 1.$$
By (4.19), (4.23) and (4.30) we have that
$$(b_6,y_5\bar{x}_5)\leq 1.$$
Therefore
$$(b_6,y_5\bar{x}_5)=1\quad\hbox{and }\quad (\bar{b}_3y_5,b_3x_5)=2$$
(4.32)
which implies that
$$\bar{b}_3y_5=b_3x_5.$$
(4.33)
In addition, (4.20) implies that
$$\bar{b}_6x_5=\bar{b}_3b_{10}.$$
(4.34)
By (4.19) we obtain that (x 5,b 3 y 5)=1. Now (4.30) implies that
$$b_3y_5=x_5+y_{10}, \quad\hbox{where } y_{10}\in B.$$
(4.35)
Now (4.33) implies that
$$\bar{b}_3y_5=b_3x_5=x+y\quad\hbox{where }x,y\in B$$
(4.36)
and (4.21) implies that
$$b_8x_5=x_5+x_{10}+y_{10}+b_{15}.$$
(4.37)
Now by (4.2) and $$\bar{b}_{3}(b_{3}x_{5})= (\bar{b}_{3}b_{3})x_{5}$$, we obtain that
$$\bar{b}_3x+\bar{b}_3y=2x_5+x_{10}+y_{10}+b_{15}.$$
(4.38)
Thus we can assume that x=z 5 and y=z 10, which imply that
$$\bar{b}_3y_5=b_3x_5=z_5+z_{10},$$
(4.39)
and one of the two following cases hold:
Case 1
$$\bar{b}_3z_5=x_5+x_{10} \quad\hbox {and}\quad \bar{b}_3z_{10}=x_5+y_{10}+b_{15}.$$
(4.40)
Case 2
$$\bar{b}_3z_5=x_5+y_{10} \quad\hbox{and}\quad \bar{b}_3z_{10}=x_5+x_{10}+b_{15}.$$
(4.41)
Thus (z 10,b 3 b 15)=1, by (4.3) we have that (b 6,b 3 b 15)=1 and together with (4.10), we obtain that
$$\begin{array}{*{20}c} {b_3 b_{15}= b_6+ z_{10}+ x_{14}+ \bar b_{15} } & {{\rm where}\,x_{14}\in B,}\\ \end{array}$$
(4.42)
$$\bar b_3 b_{10}= b_6+ z_{10}+ x_{14} ,$$
(4.43)
and by (4.34) we obtain that
$$\bar{b}_6x_5=b_6+z_{10}+x_{14}.$$
(4.44)
By (4.43) we obtain that (b 10,b 3 z 10)=1, by (4.39) we obtain that (y 5,b 3 z 10)=1, and by (4.40) and (4.41) we obtain that (b 3 z 10,b 3 z 10)=3. Thus
$$b_3z_{10}=y_5+b_{10}+x_{15} \quad\hbox{where } x_{15}\in B.$$
(4.45)
By (4.9) and (4.42), we obtain that
$$b_6b_8=\bar{b}_3+\bar{b}_{15}+b_6+z_{10}+x_{14}.$$
(4.46)
Now by (4.4) we obtain that $$(b_{6}\bar{b}_{6},b_{3}z_{10})=(\bar{b}_{3}\bar{b}_{6},\bar{b}_{6}z_{10})=(b_{8},\bar{b}_{6}z_{10})+\allowbreak(\bar{b}_{10},\bar{b}_{6}z_{10})=1+(\bar{b}_{10},\bar{b}_{6}z_{10})$$ which implies that $$(b_{6}\bar{b}_{6},b_{3}z_{10})\geq 1$$. By (4.17) we have that $$(b_{6}\bar{b}_{6},b_{6}\bar{b}_{6})=4$$. Thus either
$$b_6\bar{b}_6=1+b_8+x_{12}+x_{15}\quad\hbox{where } x_{12},x_{15}\in B$$
or
$$b_6\bar{b}_6=1+b_8+y_5+x_{22}\quad\hbox{where }y_5,x_{22}\in B.$$
Assume henceforth that $$b_{6}\bar{b}_{6}=1+b_{8}+y_{5}+x_{22}$$ and we shall derive a contradiction. Then by (4.12) we obtain that $$\bar{b}_{3}y_{5}=\bar{b}_{15}$$, and we have a contradiction to (4.30). Hence
$$b_6\bar{b}_6=1+b_8+x_{12}+x_{15}$$
(4.47)
and
$$b_3\bar{b}_{15}=\bar{b}_{10}+b_8+x_{12}+x_{15}.$$
(4.48)
By (4.1), (4.35), (4.39) and $$b_{3}(b_{3}y_{5})=b_{3}^{2}y_{5}$$, we obtain that
$$b_3y_{10}=b_6y_5.$$
(4.49)
By (4.39) we obtain that (y 5,b 3 z 5)=1, and by (4.40) and (4.41) we obtain that (b 3 z 5,b 3 z 5)=2 which implies that
$$b_3z_5=y_5+t_{10}\quad \hbox {where } t_{10}\in B.$$
(4.50)
By (4.19), (4.35), (4.39), (4.45), (4.50) and $$b_{3}(\bar{b}_{3}y_{5})=\bar{b}_{3}(b_{3}y_{5})$$, we obtain that
$$\bar{b}_3y_{10}=y_5+t_{10}+x_{15}.$$
(4.51)
By (4.4), (4.8), (4.12), (4.16), (4.19), (4.43), (4.45), (4.48) and $$\bar{b}_{3}(b_{3}b_{10})=b_{3}(\bar{b}_{3}b_{10})$$, we obtain that
$$x_{12}+\bar{b}_3x_{10}=b_3x_{14}$$
(4.52)
which implies that $$(x_{14},\bar{b}_{3}x_{12})=1$$. Now by (4.48) we obtain that
$$\bar{b}_3x_{12}=t_7+x_{14}+\bar{b}_{15}\quad\hbox{where } t_7\in B.$$
(4.53)
By (4.34) we obtain that $$3=(b_{6}x_{5},b_{6}x_{5})=(b_{6}\bar{b}_{6},x_{5}\bar{x}_{5})$$. Now by (4.47) we obtain that
$$x_5\bar{x}_5=1+b_8+x_{12}+t_4\quad\hbox{where }t_4\in B.$$
(4.54)
By (4.5), (4.19) ,(4.53), (4.54) and $$b_{3}(x_{5}\bar{x}_{5})=(b_{3}\bar{x}_{5})x_{5}$$, we obtain that
$$\bar{b}_{10}x_5+\bar{y}_5x_5=2b_3+\bar{b}_6+2b_{15}+\bar{t}_7+\bar{x}_{14}+b_3t_4.$$
(4.55)
By (4.35) we obtain that $$(b_{3},\bar{y}_{5}x_{5})=1$$, and by (4.32) we obtain that $$(b_{15},\bar{y}_{5}x_{5})=(\bar{x}_{14},\bar{y}_{5}x_{5})=0$$ which implies that
$$\bar{b}_{10}x_5=b_3+c_3+2b_{15}+\bar{x}_{14}\quad\hbox{where } c_3\in B$$
and
$$(c_3,b_3t_4)=1$$
which implies that $$(\bar{b}_{3}c_{3},\bar{b}_{3}c_{3})=2$$. By (4.2) we obtain that $$(\bar{b}_{3}c_{3},\bar{b}_{3}c_{3})=(b_{3}\bar{b}_{3},c_{3}\bar{c}_{3})=1+(b_{8},c_{3}\bar{c}_{3})$$ which implies that
$$c_3\bar{c}_3=1+b_8$$
and
$$\bar{b}_3c_3=t_4+u_5\quad\hbox{where } u_5\in B.$$
(4.56)
Now by (4.5) and $$b_{3}(c_{3}\bar{c}_{3})=(b_{3}\bar{c}_{3})c_{3}$$ we obtain that
$$c_3t_4+c_3\bar{u}_5=2b_3+\bar{b}_6+b_{15}$$
which implies that
$$c_3\bar{u}_5=b_{15}.$$
By (4.56) we obtain that $$(b_{3},c_{3}\bar{u}_{5})=1$$, which is a contradiction.

## 4.4 Case R15=x6+x9

Throughout this section we assume that R 15=x 6+x 9. Then by (4.7), (4.8) and (4.13), we have that
$$b_3 b_{10}= b_{15}+ x_6+ x_9 ,$$
(4.57)
$$b_6^2= \bar b_6+ b_{15}+ x_6+ x_9 ,$$
(4.58)
$$b_6 b_{10}= \bar b_{15}+ b_3 x_6+ b_3 x_9 .$$
(4.59)
By (4.57) we obtain that
$$(b_{10},\bar{b}_3x_6)=1$$
(4.60)
which implies that either $$2\leq (\bar{b}_{3}x_{6},\bar{b}_{3}x_{6})\leq 3$$ or $$\bar{b}_{3}x_{6}=b_{10}+2b_{4}$$.

Assume henceforth that $$\bar{b}_{3}x_{6}=b_{10}+2b_{4}$$ and we shall derive a contradiction. Then b 3 b 4=2x 6 which implies by (4.2) that $$4=(b_{3}b_{4},b_{3}b_{4})=(b_{3}\bar{b}_{3},b_{4}\bar{b}_{4})=1+(b_{8},b_{4}\bar{b}_{4})$$. Thus $$(b_{8},b_{4}\bar{b}_{4})=3$$, and we have a contradiction.

Now we have that $$2\leq (\bar{b}_{3}x_{6},\bar{b}_{3}x_{6})\leq 3$$. Assume that (b 3 x 6,b 3 x 6)=2, so we can write that
$$b_3x_6=c+d\quad \hbox{where } c,d\in B.$$
(4.61)
By (4.60) we obtain that
$$\bar{b}_3x_6=b_{10}+y_8 \quad\hbox {where }y_8\in B.$$
(4.62)
Now by (4.2), (4.57) and $$b_{3}(\bar{b}_{3}x_{6})=(b_{3}\bar{b}_{3})x_{6}$$ we obtain that
$$b_8x_6=x_9+b_{15}+b_3y_8.$$
(4.63)
By (4.1), (4.61), (4.62) and $$\bar{b}_{3}(\bar{b}_{3}x_{6})=\bar{b}_{3}^{2}x_{6}$$ we obtain that
$$\bar{b}_3b_{10}+\bar{b}_3y_8=c+d+\bar{b}_6x_6.$$
(4.64)
By (4.58) we obtain that $$(b_{6}\bar{b}_{6},b_{6}\bar{b}_{6})=4$$, by (4.4) we obtain that $$2=(b_{3}b_{6},b_{3}b_{6})=(b_{3}\bar{b}_{3},b_{6}\bar{b}_{6})$$ which implies that
$$b_6\bar{b}_6=1+b_8+s+t \quad\hbox {where s, tare real elements in B}.$$
(4.65)
By (4.61) we obtain that $$2=(b_{3}x_{6},b_{3}x_{6})=(b_{3}\bar{b}_{3},x_{6}\bar{x}_{6})$$. Now (4.2) implies that
$$(b_8,x_6\bar{x}_6)=1$$
(4.66)
and together with (4.4), we obtain that $$(b_{3}x_{6},\bar{b}_{6}x_{6})=(b_{3}b_{6},x_{6}\bar{x}_{6})=(b_{8},x_{6}\bar{x}_{6})+(b_{10},x_{6}\bar{x}_{6})=1+(b_{10},x_{6}\bar{x}_{6})$$. Thus
$$(b_3x_6,\bar{b}_6x_6)=1+(b_{10},x_6\bar{x}_6).$$
(4.67)
By (4.12) and (4.65) we obtain that
$$b_3\bar{b}_{15}=b_8+\bar{b}_{10}+s+t,$$
(4.68)
which implies that |s|,|t|≥5.

Assume henceforth that s=s 5 and we shall derive a contradiction. Then by (4.68) we obtain that $$b_{3}\bar{b}_{15}=b_{8}+\bar{b}_{10}+s_{5}+t_{22}$$ and b 3 s 5=b 15. Hence (4.2) and $$\bar{b}_{3}(b_{3}s_{5})=(b_{3}\bar{b}_{3})s_{5}$$ imply that b 8 s 5=b 8+b 10+t 22. Since b 8 and s 5 are real elements and b 10 is non-real, we have a contradiction.

Assume henceforth that s=s 6 and we shall derive a contradiction. Then by (4.68) we obtain that $$b_{3}\bar{b}_{15}=b_{8}+\bar{b}_{10}+s_{6}+t_{21}$$ and b 3 s 6=b 15+c 3 where c 3B. Now $$\bar{b}_{3}(b_{3}s_{6})=(b_{3}\bar{b}_{3})s_{6}$$ implies that $$b_{8}s_{6}=b_{8}+b_{10}+t_{21}+\bar{b}_{3}c_{3}$$. Since b 8 and s 6 are reals and b 10 is non-real, we obtain that $$(\bar{b}_{10},\bar{b}_{3}c_{3})=1$$ and we have a contradiction, which implies that
$$|s|\geq 7.$$
(4.69)
In the same way we can show that
$$|t|\geq 7.$$
(4.70)
Now (4.65) implies that $$(b_{6}\bar{b}_{6},x_{6}\bar{x}_{6})\leq 5$$. By (4.58) we obtain that $$(b_{6},\bar{b}_{6}x_{6})=1$$ and by (4.64) we obtain that $$(b_{6}\bar{b}_{6},x_{6}\bar{x}_{6})=(\bar{b}_{6}x_{6},\bar{b}_{6}x_{6})\geq 3$$. Therefore
$$3\leq (b_6\bar{b}_6,x_6\bar{x}_6)\leq 5.$$
(4.71)
Assume henceforth that $$(b_{10},x_{6}\bar{x}_{6})=1$$ and we shall derive a contradiction. Then $$x_{6}\bar{x}_{6}=1+b_{8}+b_{10}+\bar{b}_{10}+z_{7}$$, where z 7B. Now (4.65) implies that $$(\bar{b}_{6}x_{6},\bar{b}_{6}x_{6})\leq 3$$ and together with (4.71) we obtain that $$(\bar{b}_{6}x_{6},\bar{b}_{6}x_{6})=3$$, and by (4.61), (4.67) and $$(b_{10},x_{6}\bar{x}_{6})=1$$ we obtain that $$\bar{b}_{6}x_{6}=b_{6}+c+d$$, which is a contradiction. Thus (4.67) implies that
$$(b_{10},x_6\bar{x}_6)=0$$
(4.72)
and
$$(b_3x_6,\bar{b}_6x_6)=1.$$
(4.73)
Now we can assume that
$$(c,\bar{b}_6x_6)=1.$$
(4.74)
In addition, (4.1) and (4.62) imply that $$(\bar{b}_{3}y_{8},b_{3}x_{6})=(b_{3}^{2},y_{8}\bar{x}_{6})=(\bar{b}_{3},y_{8}\bar{x}_{6})+(b_{6},y_{8}\bar{x}_{6})=(y_{8},\bar{b}_{3} x_{6})+(y_{8},b_{6}x_{6})=1+(y_{8},b_{6}x_{6})$$. By (4.62) and (4.73), we obtain that $$1=(b_{3}x_{6},\bar{b}_{6}x_{6})=(\bar{b}_{3}x_{6},b_{6}x_{6})=(b_{10},b_{6}x_{6})+(y_{8},b_{6}x_{6})$$ which implies that either $$(\bar{b}_{3}y_{8},b_{3}x_{6})=1$$ or $$(\bar{b}_{3}y_{8},b_{3}x_{6})=2$$. Assume that $$(\bar{b}_{3}y_{8},b_{3}x_{6})=1$$. Then (4.4), (4.64), (4.74) and (b 3 b 10,b 3 b 10)=3 imply that $$\bar{b}_{3}b_{10}=b_{6}+c+d$$ which is a contradiction, and we obtain that $$(\bar{b}_{3}y_{8},b_{3}x_{6})=2$$. Now we have that either $$(c,\bar{b}_{3}y_{8})=2$$ or $$(c,\bar{b}_{3}y_{8})=(d,\bar{b}_{3}y_{8})=1$$, which implies that (b 3 y 8,b 3 y 8)≥3. By (4.11), (4.66) and (4.72), we obtain that $$(b_{8}x_{6},b_{8}x_{6})=(b_{8}^{2},x_{6}\bar{x}_{6})=1+(b_{6}\bar{b}_{6},x_{6}\bar{x}_{6})$$. Now (4.71) implies that (b 8 x 6,b 8 x 6)≤6 together with (4.63) and since (b 3 y 8,b 3 y 8)≥3, we have that
$$(x_9,b_3y_8)=(b_{15},b_3y_8)=0,$$
(4.75)
which implies that
$$(b_8x_6,b_8x_6)=2+(b_3y_8,b_3y_8).$$
(4.76)
Therefore (b 3 y 8,b 3 y 8)≤4. If $$(c,\bar{b}_{3}y_{8})=2$$ then $$\bar{b}_{3}y_{8}=2c_{12}$$ and $$(\bar{b}_{6}x_{6},\bar{b}_{6}x_{6})=5$$. Now (4.64) implies that $$\bar{b}_{3}b_{10}=b_{6}+d_{6}+f_{18}$$ where f 18B and $$\bar{b}_{6}x_{6}=b_{6}+c_{12}+f_{18}$$, which is a contradiction.
Now we have that $$(c,\bar{b}_{3}y_{8})=(d,\bar{b}_{3}y_{8})=1$$ which implies that
$$\bar{b}_3y_8=c+d+e_6 \quad\hbox{where } e_6\in B.$$
(4.77)
Now (4.64) and (4.74) imply that
$$\begin{array}{*{20}c} {\bar b_3 b_{10}= b_6+ c + f} & {{\rm where}\,f \in B,}\\ \end{array}$$
(4.78)
$$\bar b_6 x_6= b_6+ e_6+ c + f,$$
(4.79)
and by (4.10) we obtain that
$$b_3b_{15}=\bar{b}_{15}+b_6+c+f.$$
(4.80)
By (4.2), (4.61), (4.63) and $$\bar{b}_{3}(b_{3}x_{6})=(b_{3}\bar{b}_{3})x_{6}$$, we obtain that
$$\bar{b}_3c+\bar{b}_3d=x_6+x_9+b_{15}+b_3y_8.$$
(4.81)
Now we can assume that $$(b_{15},\bar{b}_{3}c)=1$$. By (4.61) we obtain that $$(x_{6},\bar{b}_{3}c)=(x_{6},\bar{b}_{3}d)=1$$ and by (4.77) we obtain that (b 3 y 8,b 3 y 8)=3 which implies by (4.75) and (4.81) that 2≤(b 3 c,b 3 c)≤4. By (4.77) and (4.78), we obtain that (b 10,b 3 c)=(y 8,b 3 c)=1 which implies that (b 3 c,b 3 c)≠2.
Therefore
$$3\leq (b_3c,b_3c)\leq 4.$$
(4.82)
By (4.77) we obtain that (b 3 y 8,b 3 y 8)=3 and by (4.62), we obtain that $$(y_{8},\allowbreak\bar{b}_{3}x_{6})=\nobreak 1$$ which implies that
$$b_3y_8=x_6+\alpha+\beta \quad\hbox {where } \alpha, \beta\in B.$$
(4.83)
Now by (4.81) we obtain that
$$\bar{b}_3c+\bar{b}_3d=2x_6+x_9+b_{15}+\alpha+\beta.$$
(4.84)
Assume that (b 3 c,b 3 c)=3. Then by (4.61) and (4.80) we obtain that
$$\bar{b}_3c=x_6+b_{15}+g, \quad\hbox {where } g\in B$$
(4.85)
which implies that |c|≥9, and by (4.84) we obtain that $$g+\bar{b}_{3}d=x_{6}+x_{9}+\alpha+\beta$$. If g=x 9 then by (4.85) we obtain that |c|=10, and if gx 9 then we obtain that either g=α or g=β. If α=3=α 3, then (4.5) and (4.83) imply that α=b 3, y 8=b 8 and $$x_{6}=\bar{b}_{6}$$ which implies by (4.57) that $$b_{3}b_{10}=b_{15}+\bar{b}_{6}+x_{9}$$. Thus $$(\bar{b}_{10},b_{3}b_{6})=1$$, and we have a contradiction to (4.4). In the same way, we can show that |β|≠3 which implies by (4.83) that
$$4\leq |\alpha|,|\beta|\leq 14.$$
(4.86)
Now (4.85) implies that |g|≤12 and |c|≤11. Therefore
$$9\leq |c|\leq 11.$$
(4.87)
Assume that (b 3 c,b 3 c)=4. Then by (4.62) and by (4.68), we obtain that
$$\bar{b}_3c=x_6+b_{15}+g+h\quad\hbox{where } g,h\in B.$$
(4.88)
Now by (4.81) and (4.83) we obtain that $$\bar{b}_{3}c+\bar{b}_{3}d=2x_{6}+x_{9}+b_{15}+\alpha+\beta$$, which implies that
$$g+h+\bar{b}_3d=x_6+x_9+\alpha+\beta.$$
(4.89)
If $$(x_{9},\bar{b}_{3}d)=1$$, then g+h=α+β which implies by (4.83) and (4.88) that |c|=13. If $$(x_{9},\bar{b}_{3}d)=0$$ then we can assume that g+h=x 9+α. Now from 4≤|α|≤14 and by (4.88) we obtain that
$$12\leq |c|\leq 14.$$
(4.90)
From (4.61), (4.78), (4.87) and (4.90) we have seven cases (see Table 4.2).
Table 4.2

Splitting to seven cases

 Case 1 c=c 9 d=d 9 f=f 15 (b 3 c 9,b 3 c 9)=3 Case 2 c=c 10 d=d 8 f=f 14 (b 3 c 10,b 3 c 10)=3 Case 3 c=c 11 d=d 7 f=f 13 (b 3 c 11,b 3 c 11)=3 Case 4 c=c 12 d=d 6 f=f 12 (b 3 c 12,b 3 c 12)=4 Case 5 c=c 13 d=d 5 f=f 11 (b 3 c 13,b 3 c 13)=4 Case 6 c=c 14 d=d 4 f=f 10 (b 3 c 14,b 3 c 14)=4 Case 7 (b 3 x 6,b 3 x 6)=3
In the following seven sections we derive a contradiction for each case in the above table, and in the following three sections we assume that
$$(b_3c,b_3c)=3,$$
(4.91)
so (4.87) implies that 9≤|c|≤11.

### 4.4.1 Case c=c9

In this section we assume that c=c 9. Then (4.77), (4.78) and (4.91) imply that b 3 c 9=y 8+b 10+q 9 where q 9B which implies that
$$(c_9,\bar{b}_3q_9)=1.$$
(4.92)
By (4.85) we obtain that $$\bar{b}_{3}c_{9}=x_{6}+b_{15}+g_{6}$$. Now by (4.61), (4.77), (4.78), (4.80) and $$\bar{b}_{3}(b_{3}c_{9})=b_{3}(\bar{b}_{3}c_{9})$$, we obtain that $$\bar{b}_{15}+b_{3}g_{6}=e_{6}+\bar{b}_{3}q_{9}$$ which implies that $$(\bar{b}_{15},\bar{b}_{3}h_{9})=1$$, and together with (4.92) we obtain that $$\bar{b}_{3}h_{9}=c_{9}+\bar{b}_{15}+c_{3}$$ where c 3B, b 3 c 3=h 9 and (c 3,b 3 g 6)=1. Since b 3 c 3=h 9 we obtain that (b 3 c 3,b 3 c 3)=1, which is a contradiction to (c 3,b 3 g 6)=1.

### 4.4.2 Case c=c10

In this section we assume that c=c 10. Then (4.77), (4.78) and (4.91) imply that
$$b_3c_{10}=y_8+b_{10}+q_{12}\quad\hbox{where } q_{12}\in B.$$
(4.93)
By (4.85) we obtain that
$$\bar{b}_3c_{10}=x_6+b_{15}+g_9.$$
(4.94)
Now by (4.61), (4.77), (4.78), (4.80) and $$\bar{b}_{3}(b_{3}c_{10})=b_{3}(\bar{b}_{3}c_{10})$$, we obtain that $$\bar{b}_{15}+b_{3}g_{9}=e_{6}+\bar{b}_{3}q_{12}$$ which implies that
$$(q_{12},b_3\bar{b}_{15})=1,$$
(4.95)
and (4.12) implies that q 12 is a real element. By (4.62), (4.77), (4.83), (4.93) and $$b_{3}(\bar{b}_{3}y_{8})=\bar{b}_{3}(b_{3}y_{8})$$ we obtain that $$q_{12}+b_{3}d_{8}+b_{3}e_{6}=\bar{b}_{3}\alpha+\bar{b}_{3}\beta$$. Now we can assume that $$(q_{12},\bar{b}_{3}\alpha)=1$$ and by (4.83) we obtain that $$(y_{8},\bar{b}_{3}\alpha)=1$$ which implies that |α|≥8, and since q 12 is a real element we obtain that
$$(\bar{\alpha},\bar{b}_3 q_{12})=1.$$
(4.96)

Assume henceforth that $$\alpha=\bar{c}_{10}$$ and we shall derive a contradiction. Then by (4.81), (4.83) and (4.94) we obtain that $$(\bar{c}_{10},\bar{b}_{3}d_{8})=1$$, and we have a contradiction to (4.94). By (4.86) we obtain that αb 15. Now by (4.93), (4.95) and (4.96) we obtain that $$(c_{10},\bar{b}_{3}q_{12})=1$$, $$(\bar{\alpha},\bar{b}_{3}q_{12})=1$$ and $$(\bar{b}_{15},\bar{b}_{3} q_{12})=1$$ which imply that either |α|≤7 or |α|=11. Now |α|≥8 implies that α=α 11 and by (4.83) we obtain that β=β 7.

Now we have that
$$b_3q_{12}=\bar{c}_{10}+b_{15}+\alpha_{11}.$$
(4.97)
By (4.5), (4.8) and (4.95) we obtain that $$\bar{b}_{3}b_{15}=b_{8}+b_{10}+q_{12}+q_{15}$$ where q 15B. Hence (4.2), (4.93) and $$\bar{b}_{3}(b_{3}q_{12})=(b_{3}\bar{b}_{3})q_{12}$$ imply that
$$b_8q_{12}=\bar{y}_8+\bar{b}_{10}+q_{12}+b_8+b_{10}+q_{15}+\bar{b}_3\alpha_{11}.$$
(4.98)
By (4.62), (4.77), (4.83), (4.93) and $$b_{3}(\bar{b}_{3}y_{8})=\bar{b}_{3}(b_{3}y_{8})$$ we obtain that
$$\bar{b}_3\alpha_{11}+\bar{b}_3\beta_7=q_{12}+b_3d_8+b_3e_6$$
(4.99)
and
$$(y_8,\bar{b}_3\alpha_{11})=(y_8,b_3e_6)=1.$$
(4.100)
If $$(\bar{b}_{3}\alpha_{11},b_{3}e_{6})=1$$, then by (4.97) we obtain that $$\bar{b}_{3}\alpha_{11}=q_{12}+y_{8}+\varSigma _{13}$$ where Σ 13∈ℕB and b 3 d 8=y 8+Σ 13+ν 3 where ν 3B which implies that $$(\nu_{3},\allowbreak\bar{b}_{3}\beta_{7})=\nobreak 1$$ and we have a contradiction. Now we have that $$(\bar{b}_{3}\alpha_{11},b_{3}e_{6})\geq 2$$.

If b 3 e 6=y 8+r 10 where r 10B, then $$\bar{b}_{3}\alpha_{11}=c_{3}+y_{8}+r_{10}+q_{12}$$ and we have a contradiction.

If b 3 e 6=y 8+r 4+r 6 where r 4,r 6B, then $$(r_{6},\bar{b}_{3}\alpha_{11})=1$$ and (4.98) implies that r 6 is a real element, and we have that $$(\bar{e}_{6},b_{3}r_{6})=(\alpha_{11},b_{3}r_{6})=1$$, which is a contradiction.

If b 3 e 6=y 8+r 5+γ 5 where r 5,γ 5B then we have that $$(\bar{e}_{6},b_{3}r_{5})=(\alpha_{11},b_{3}r_{5})=1$$ and we have a contradiction.

### 4.4.3 Case c=c11

In this section we assume that c=c 11. Then (4.77), (4.78) and (4.91) imply that
$$b_3c_{11}=y_8+b_{10}+q_{15}\quad\hbox{where } q_{15}\in B.$$
(4.101)
By (4.85) we obtain that
$$\bar{b}_3c_{11}=x_6+b_{15}+g_{12}.$$
(4.102)
Now by (4.61), (4.77), (4.78), (4.80) and $$\bar{b}_{3}(b_{3}c_{11})=b_{3}(\bar{b}_{3}c_{11})$$, we obtain that
$$\bar{b}_{15}+b_3g_{12}=e_6+\bar{b}_3q_{15}$$
(4.103)
which implies that
$$(q_{15},b_3\bar{b}_{15})=1.$$
(4.104)
Now (4.12) implies that q 15 is a real element, and we obtain that
$$(b_{15},b_3q_{15})=1.$$
(4.105)
By (4.81), (4.83) and (4.102) we obtain that
$$b_3y_8=x_6+\alpha_6+g_{12}.$$
(4.106)
Thus β=g 12.
By (4.62), (4.77), (4.101) and $$b_{3}(\bar{b}_{3}y_{8})=\bar{b}_{3}(b_{3}y_{8})$$, we obtain that
$$\bar{b}_3\alpha_{6}+\bar{b}_3g_{12}=q_{15}+b_3d_7+b_3e_6.$$
(4.107)
Now by (4.106) we obtain that $$(y_{8},\bar{b}_{3}\alpha_{6})=1$$. Hence $$(q_{15},\bar{b}_{3}\alpha_{6})=0$$, which implies that
$$1=(q_{15},\bar{b}_3g_{12})=(b_3q_{15},g_{12}).$$
(4.108)
Since q 15 is a real element, (4.101) implies that
$$(\bar{c}_{11},b_3q_{15})=1.$$
Now by (4.105) and (4.108) we obtain that
$$b_3q_{15}=\bar{c}_{11}+g_{12}+b_{15}+z_7 \quad\hbox{where } z_7\in B$$
(4.109)
which implies that
$$\bar{b}_3z_7=q_{15}+z_6 \quad\hbox{where } z_6\in B.$$
(4.110)
By (4.103), (4.109) and since q 15 is a real element, we have that
$$b_3g_{12}=e_6+\bar{z}_7+c_{11}+\bar{g}_{12},$$
which implies that
$$b_3z_7=\bar{g}_{12}+r_9,\quad\hbox{where } r_9\in B.$$
Now $$b_{3}(\bar{b}_{3}z_{7})=\bar{b}_{3}(b_{3}z_{7})$$ implies that
$$b_{15}+b_3z_6=\bar{e}_6+\bar{b}_3r_9.$$
Hence $$1=(b_{15},\bar{b}_{3}r_{9})=(r_{9},b_{3}b_{15})$$ and we have a contradiction to (4.80).
In the following three sections, we assume that
$$(b_3c,b_3c)=4,$$
(4.111)
so (4.90) implies that 12≤|c|≤14.

### 4.4.4 Case c=c12

In this section we assume that c=c 12. Then by (4.61), (4.80), (4.81) and (4.111), we obtain that
$$\bar{b}_3c_{12}=x_6+x_9+b_{15}+g_6\quad\hbox{where } g_6\in B$$
(4.112)
and
$$b_3y_8=x_6+g_6+\beta_{12}\quad\hbox{where}\beta_{12}\in B.$$
(4.113)
Assume henceforth that (b 3 g 6,b 3 g 6)=3 and we shall derive a contradiction. Then (4.112) implies that b 3 g 6=c 12+v 3+w 3 and $$\bar{b}_{3}g_{6}=y_{8}+\nu+\mu$$ where v 3,w 3,ν,μB. Now by (4.112), (4.113) and $$\bar{b}_{3}(b_{3}g_{6})=b_{3}(\bar{b}_{3}g_{6})$$, we obtain that $$x_{9}+b_{15}+\bar{b}_{3}v_{3}+\bar{b}_{3}w_{3}=\beta_{12}+b_{3}\nu+b_{3}\mu$$, which is a contradiction. Now we have that (b 3 g 6,b 3 g 6)=2, by (4.112) and (4.113) we obtain that
$$b_3g_6=c_{12}+v_6 \quad\hbox {where } v_6\in B$$
(4.114)
and
$$\bar{b}_3g_6=y_8+v_{10}\quad \hbox{where }v_{10}\in B.$$
(4.115)
Now by (4.112), (4.113) and $$\bar{b}_{3}(b_{3}g_{6})=b_{3}(\bar{b}_{3}g_{6})$$ we obtain that
$$x_9+b_{15}+\bar{b}_3v_6=\beta_{12}+b_3v_{10},$$
which implies that
$$(v_{10},\bar{b}_3b_{15})=1.$$
(4.116)
In addition, we have that $$(b_{10},\bar{b}_{3}b_{15})=(b_{8},\bar{b}_{3}b_{15})=1$$ which implies that
$$\bar{b}_3b_{15}=b_8+b_{10}+v_{10}+v_{17} \quad\hbox{where } v_{17}\in B.$$
(4.117)
Now by (4.12) we obtain that v 10 and v 17 are real elements. By (4.77), (4.78) and (4.111) we obtain that
$$b_3c_{12}=y_8+b_{10}+q+u \quad\hbox{where } q,u\in B.$$
(4.118)
Now by (4.61), (4.77), (4.78), (4.80), (4.112), (4.114) and $$\bar{b}_{3}(b_{3}c_{12})=b_{3}(\bar{b}_{3}c_{12})$$, we obtain that
$$e_6+\bar{b}_3q+\bar{b}_3u=\bar{b}_{15}+c_{12}+v_6+b_3x_9.$$
Now we can assume that $$(q,b_{3}\bar{b}_{15})=1$$; in addition qy 8,b 10, and by (4.118) we obtain that |q|≠17. Thus qv 17 which implies that q=v 10. Now since v 10 is a real element, (4.118) implies $$(\bar{c}_{12},b_{3}v_{10})=1$$ and by (4.115) we obtain that (g 6,b 3 v 10)=1, which is a contradiction to (4.116).

### 4.4.5 Case c=c13

In this section we assume that c=c 13. Then by (4.77) we obtain that
$$\bar{b}_3y_8=d_5+e_6+c_{13}.$$
(4.119)
By (4.61) and (4.81) we obtain that either
$$\bar{b}_3d_5=x_6+x_9$$
(4.120)
or
$$\bar{b}_3d_5=x_6+z_9 \quad\hbox {where }z_9\in B.$$
(4.121)
Now we have that (b 3 d 5,b 3 d 5)=2 and together with (4.119), we obtain that
$$b_3d_5=y_8+z_7 \quad\hbox{where } z_7\in B.$$
(4.122)
Assume that $$\bar{b}_{3}d_{5}=x_{6}+z_{9}$$. Then by (4.81) we obtain that
$$\bar{b}_3c_{13}=x_6+x_9+b_{15}+r_9\quad\hbox{where } r_9\in B$$
(4.123)
and
$$b_3y_8=x_6+z_9+r_9.$$
(4.124)
Now by (4.1), (4.121), (4.122) and $$b_{3}(b_{3}d_{5})=b_{3}^{2}d_{5}$$, we obtain that r 9+b 3 z 7=b 6 d 5. By (4.65), (4.69) and (4.70), we obtain that (b 6 d 5,b 6 d 5)≤3 which implies that
$$(r_9,b_3z_7)=0.$$
(4.125)
By (4.77), (4.78) and (4.123), we obtain that
$$b_3c_{13}=y_8+b_{10}+q+u \quad\hbox{where } q,u\in B.$$
(4.126)
Now by (4.62), (4.119), (4.122), (4.124), and $$b_{3}(\bar{b}_{3}y_{8})=\bar{b}_{3}(b_{3}y_{8})$$ we obtain that
$$q+u+y_8+z_7+b_3e_6=\bar{b}_3z_9+\bar{b}_3r_9,$$
(4.127)
and together with (4.125), we obtain that
$$(z_9,b_3z_7)=1.$$
(4.128)
By (4.126) we obtain that
$$|q|+|u|=21,$$
(4.129)
and by ( 4.124 ) we obtain that $$(y_{8},\bar{b}_{3}z_{9})=(y_{8},\bar{b}_{3}r_{9})=1$$. Now we can assume by (4.127) that
$$(q,\bar{b}_3r_9)=(u,\bar{b}_3z_9)=1,$$
(4.130)
and together with (4.128) we obtain that |u|≤12.
By (4.80) we obtain that $$b_{3}b_{15}=\bar{b}_{15}+b_{6}+c_{13}+f_{11}$$, and by (4.78) we obtain that $$\bar{b}_{3}b_{10}=b_{6}+c_{13}+f_{11}$$. Now by (4.61), (4.119), (4.123), (4.126) and $$b_{3}(\bar{b}_{3}c_{13})=\bar{b}_{3}(b_{3}c_{13})$$, we obtain that
$$\bar{b}_{15}+b_3x_9+b_3r_9=e_6+\bar{b}_3q+\bar{b}_3u$$
which implies that either $$(\bar{b}_{15},\bar{b}_{3}q)=1$$ or $$(\bar{b}_{15},\bar{b}_{3}u)=1$$.

Assume henceforth that $$(u,b_{3}\bar{b}_{15})=1$$ and we shall derive a contradiction. Then by (4.12) we obtain that u is a real element. By (4.130) we obtain that (z 9,b 3 u)=1, by (4.126) we obtain that $$(\bar{c}_{13},b_{3}u)=1$$, and by the assumption we obtain that (b 15,b 3 u)=1 which implies that |u|≥14, and we have a contradiction to |u|≤12.

Now we obtain that
$$(q,b_3\bar{b}_{15})=1$$
(4.131)
which implies by (4.12) that q is a real element. By (4.130) we obtain that (r 9,b 3 q)=1, by (4.126) we obtain that $$(\bar{c}_{13},b_{3}q)=1$$, and together with (b 15,b 3 q)=1 we obtain that |q|≥14.
Since $$(c_{13},\bar{b}_{3}u)=1$$ we obtain that |u|≥6. Now (4.129) implies that |q|≤15, and we obtain that
$$14\leq |q|\leq 15$$
(4.132)
and
$$6\leq |u|\leq 7.$$
(4.133)
By (4.123), (4.124) and (4.130) we obtain that
$$\bar{b}_3r_9=y_8+q+\lambda \quad\hbox{where }\lambda\in B \hbox{ and } 4\leq\lambda\leq 5$$
and
$$b_3r_9=c_{13}+m+n \quad\hbox{where } m,n\in B.$$
Now by (4.123), (4.124) and $$b_{3}(\bar{b}_{3}r_{9})=\bar{b}_{3}(b_{3}r_{9})$$, we obtain that
$$z_9+b_3q+b_3\lambda=x_9+b_{15}+\bar{b}_3m+\bar{b}_3n.$$
Since (r 9,b 3 λ)=1 and 4≤λ≤5, we obtain that (x 9,b 3 λ)=0 which implies that either x 9=z 9 or (x 9,b 3 q)=1. If x 9=z 9 then by (4.124), (4.128) and (4.57), we obtain that $$(y_{8},\bar{b}_{3}x_{9})=(z_{7},\bar{b}_{3}x_{9})=(b_{10},\bar{b}_{3}x_{9})=1$$ and we have a contradiction.

Now we have that (x 9,b 3 q)=1. By (4.126), (4.130) and (4.131) we obtain that $$(b_{15},b_{3}q)=(\bar{c}_{13},b_{3}q)=(r_{9},b_{3}q)=1$$, and we have a contradiction to (x 9,b 3 q)=1.

Now we have that
$$\bar{b}_3d_5=x_6+x_9.$$
(4.134)
By (4.81) and (4.83) we obtain that
$$\bar{b}_3c_{13}=x_6+b_{15}+\alpha+\beta$$
(4.135)
and
$$b_3y_8=x_6+\alpha+\beta.$$
(4.136)
Now by (4.62), (4.119), (4.122), (4.126) and $$\bar{b}_{3}(b_{3}y_{8})=b_{3}(\bar{b}_{3}y_{8})$$ we obtain that
$$y_8+z_7+q+u+b_3e_6=\bar{b}_3\alpha+\bar{b}_3\beta.$$
Now we can assume that
$$(\alpha,b_3z_7)=1.$$
(4.137)
By (4.1), (4.122), (4.134), (4.136) and $$b_{3}(b_{3}d_{5})=b_{3}^{2}d_{5}$$, we obtain that
$$x_9+b_6d_5=\alpha+\beta+b_3z_7.$$
By (4.65), (4.69) and (4.70) we obtain that (b 6 d 5,b 6 d 5)≤3 which implies that x 9=α. By (4.57), (4.136) and (4.137) we obtain that $$(z_{7},\bar{b}_{3}x_{9})=(b_{10},\bar{b}_{3}x_{9})=(y_{8},\bar{b}_{3}x_{9})=1$$, which is a contradiction.

### 4.4.6 Case c=c14

In this section we assume that c=c 14. Then by (4.77) we obtain that
$$\bar{b}_3y_8=d_4+e_6+c_{14}$$
(4.138)
which implies
$$b_3d_4=y_8+y_4\quad\hbox{where } y_4\in B,$$
(4.139)
and by (4.78) we obtain that
$$b_3c_{14}=b_{10}+y_8+q+u\quad\hbox{where } q,u\in B.$$
(4.140)
Now by (4.61) we obtain that
$$\bar{b}_3d_4=x_6+y_6\quad\hbox{where } y_6\in B$$
(4.141)
and by (4.81) we obtain that
$$b_3y_8=x_6+y_6+y_{12}\quad\hbox{where } y_{12}\in B.$$
(4.142)
Now by (4.62) and $$\bar{b}_{3}(b_{3}y_{8})=b_{3}(\bar{b}_{3}y_{8})$$ we obtain that
$$\bar{b}_3y_6+\bar{b}_3y_{12}=q+u+y_8+y_4+b_3e_6.$$
(4.143)
By (4.139) we obtain that
$$(b_3y_4,b_3y_4)\neq 1$$
(4.144)
which implies that
$$(y_4,\bar{b}_3y_{12})=0,$$
(4.145)
and by (4.143) we obtain that
$$(y_4,\bar{b}_3y_6)=1.$$
(4.146)
By (4.140) we obtain that |q|+|u|=24, and by (4.142) we obtain that $$(y_{8},\allowbreak \bar{b}_{3}y_{12})=\nobreak 1$$.

Assume henceforth that $$(u,\bar{b}_{3}y_{12})=(q,\bar{b}_{3}y_{12})=1$$ and we shall derive a contradiction. Then $$\bar{b}_{3}y_{12}=q+u+y_{8}+z_{4}$$ where z 4B which implies that b 3 z 4=y 12 and by (4.143) and (4.145) we obtain that (z 4,b 3 e 6)=1, which is a contradiction to b 3 z 4=y 12.

Now we can assume that $$(u,\bar{b}_{3}y_{12})=1$$ and $$(q,\bar{b}_{3}y_{6})=1$$. By (4.142) and (4.146) we obtain that
$$\bar{b}_3y_6=y_4+y_8+q_6,$$
and by (4.140) we obtain that
$$b_3c_{14}=b_{10}+y_8+q_6+u_{18}$$
which implies that
$$\bar{b}_3q_6=c_{14}+z_4\quad\hbox{where } z_4\in B$$
and
$$b_3q_6=y_6+z_{12}\quad\hbox{where } z_{12}\in B.$$
Now by $$b_{3}(\bar{b}_{3}q_{6})=\bar{b}_{3}(b_{3}q_{6})$$ we obtain that
$$b_{10}+u_{18}+b_3z_4=e_4+\bar{b}_3z_{12},$$
and we have a contradiction to (4.57).

### 4.4.7 Case (b3x6,b3x6)=3

In this section we assume that (b 3 x 6,b 3 x 6)=3 Then
$$b_3x_6=c+d+e\quad\hbox{where } c,d,e\in B$$
(4.147)
and by (4.57) we obtain that
$$\bar{b}_3x_6=b_{10}+v+w\quad\hbox{where } v,w\in B.$$
(4.148)
Now by (4.57) and $$b_{3}(\bar{b}_{3}x_{6})=\bar{b}_{3}(b_{3}x_{6})$$ we obtain that
$$\bar{b}_3c+\bar{b}_3d+\bar{b}_3e=b_{15}+x_6+x_9+b_3v+b_3w.$$
(4.149)
Now we can assume
$$(b_{15},\bar{b}_3c)=1$$
(4.150)
and by (4.147) we obtain that $$(x_{6},\bar{b}_{3}c)=1$$ which implies that |c|≥7, and by (4.147) we obtain that |c|≤12. Now we have that
$$7\leq |c|\leq 12.$$
(4.151)
By (4.4), (4.10) and (4.150) we obtain that
$$\bar{b}_3b_{10}=b_6+c+f\quad\hbox{where } f\in B$$
(4.152)
and
$$b_3b_{15}=\bar{b}_{15}+b_6+c+f.$$
(4.153)
Assume henceforth that v=v 3 and we shall derive a contradiction. Then by (4.148) we obtain that
$$\begin{array}{*{20}c} {\bar b_3 x_6= b_{10}+ v_3+ w_5 ,}\\ {\begin{array}{*{20}c} {b_3 v_3= x_6+ z_3 } & {{\rm where}\,z_3\in B}\\ \end{array}}\\ \end{array}$$
and
$$\bar{b}_3v_3=\alpha+\beta\quad\hbox{where }\alpha,\beta\in B.$$
(4.154)
Now by $$b_{3}(\bar{b}_{3}v_{3})=\bar{b}_{3}(b_{3}v_{3})$$ we obtain that
$$b_3\alpha+b_3\beta=b_{10}+v_3+w_5+\bar{b}_3z_3$$
and we can assume that
$$(\alpha,\bar{b}_3b_{10})=1.$$
(4.155)
By (4.4) and (4.154) we obtain that αb 6 and |α|≤6, so (4.151) and (4.152) imply that αc and αf. Now we have a contradiction to (4.152) and (4.155).
Hence
$$\bar{b}_3x_6=b_{10}+v_4+w_4$$
(4.156)
which implies that
$$(x_6,b_3b_{10})=(x_6,b_3v_4)=1,$$
(4.157)
and we obtain
$$(b_3v_4,b_3v_4)=2$$
(4.158)
and $$1=(b_{3}b_{10},b_{3}v_{4})=(\bar{b}_{3}b_{10},\bar{b}_{3}v_{4})$$. By (4.4) we obtain that $$(b_{6},\bar{b}_{3}v_{4})=0$$. By (4.151) and (4.152) we obtain that 12≤f≤17, and together with (b 3 v 4,b 3 v 4)=2 we obtain that $$(f,\bar{b}_{3}v_{4})=0$$. Now we have that
$$(c,\bar{b}_3v_4)=1$$
(4.159)
which implies that |c|≤9, and together with (4.151) we obtain that
$$7\leq c\leq 9.$$
If c=c 7 then by (4.147) and (4.153) we obtain that $$\bar{b}_{3}c_{7}=b_{15}+x_{6}$$, and by (4.152) and (4.159) we obtain that b 3 c 7=v 4+b 10, which is a contradiction.

If c=c 8 then by (4.147) and (4.153), we obtain that $$\bar{b}_{3}c_{8}=b_{15}+x_{6}+r_{3}$$ where r 3B. Now we have that $$r_{3}=\bar{b}_{3}$$ and b 8=c 8, since b 15 is a non-real element we have a contradiction to (4.5).

Now we have that c=c 9 then by (4.147) and (4.153), we obtain that
$$\bar{b}_3c_9=b_{15}+x_6+\varSigma _6 \quad\hbox{where} \varSigma _6\in\mathbb{N}B,$$
(4.160)
and by (4.152) and (4.159) we obtain that
$$b_3c_9=v_4+b_{10}+\varSigma _{13} \quad\hbox{where }\varSigma _{13}\in\mathbb{N}B.$$
(4.161)
By (4.157), (4.158) and (4.159) we obtain that
$$b_3v_4=x_6+m_6\quad\hbox{where }m_6\in B$$
and
$$\bar{b}_3v_4=c_9+m_3\quad\hbox{where } m_3\in B.$$
Now by (4.156) and $$\bar{b}_{3}(b_{3}v_{4})=b_{3}(\bar{b}_{3}v_{4})$$ we obtain that
$$w_4+\bar{b}_3m_6=\varSigma _{13}+b_3m_3.$$
Now (v 4,b 3 m 3)=1 implies that (w 4,b 3 m 3)=0, so (w 4,Σ 13)=1. By (4.161) we obtain that
$$b_3c_9=v_4+b_{10}+w_4+w_9\quad\hbox{where } w_9\in B,$$
and by (4.160) we obtain that
$$\bar{b}_3c_9=b_{15}+x_6+l_3+n_3\quad\hbox{where } l_3,n_3\in B.$$
Now by (4.149) we obtain that b 3 w 4=x 6+l 3+n 3, which is a contradiction to (b 3 w 4,b 3 w 4)≤2.

## 4.5 Case R15=x7+x8

Throughout this section we assume that R 15=x 7+x 8. Then by (4.7) and (4.8) we have that
$$b_3 b_{10}= b_{15}+ x_7+ x_8 ,$$
(4.162)
$$b_6^2= \bar b_6+ b_{15}+ x_7+ x_8 .$$
(4.163)
Now $$(b_{6}\bar{b}_{6},b_{6}\bar{b}_{6})=4$$ and by (4.4) and (4.2), we obtain that
$$b_6\bar{b}_6=1+b_8+s+t\quad\hbox{where s, t arereal elements in B}$$
(4.164)
and by (4.12), we obtain that
$$\bar{b}_3b_{15}=b_8+b_{10}+s+t.$$
(4.165)
Assume henceforth that s=s 21 and we shall derive a contradiction. Then by (4.165) t=t 6 and b 3 t 6=b 15+c 3 where c 3B. Now $$\bar{b}_{3}(b_{3}t_{6})=(b_{3}\bar{b}_{3})t_{6}$$ implies that $$b_{8}t_{6}=b_{8}+b_{10}+s_{21}+\bar{b}_{3}c_{3}$$. Since b 8 and t 6 are real elements and b 10 is a non-real element, we obtain that $$(\bar{b}_{10},\bar{b}_{3}c_{3})=1$$ and we have a contradiction. Now we have that
$$|s|\neq 21.$$
(4.166)
By (4.13) we obtain that
$$b_6b_{10}=\bar{b}_{15}+b_3x_7+b_3x_8.$$
(4.167)
By (4.162) we obtain that
$$(b_{10},\bar{b}_3x_7)=1$$
(4.168)
which implies that $$2\leq (\bar{b}_{3}x_{7},\bar{b}_{3}x_{7})\leq 3$$. Assume that (b 3 x 7,b 3 x 7)=2. Then we obtain that
$$b_3x_7=c+d \quad\hbox{where } c,d\in B,$$
(4.169)
and by (4.162) we obtain that
$$\bar{b}_3x_7=b_{10}+y_{11}\quad\hbox{where }y_{11}\in B.$$
(4.170)
By (4.1) and (4.162) we obtain that $$(b_{3}x_{7},\bar{b}_{3}b_{10})=(b_{3}^{2},\bar{x}_{7}b_{10})=(\bar{b}_{3},\bar{x}_{7}b_{10})+(b_{6},\bar{x}_{7}b_{10})=(x_{7},b_{3}b_{10})+(b_{6},\bar{x}_{7}b_{10})=1+(b_{6},\bar{x}_{7}b_{10})$$ which implies that $$(b_{3}x_{7},\bar{b}_{3}b_{10})\geq 1$$. By (4.4) we obtain that $$(b_{6},\bar{b}_{3}b_{10})=1$$ which implies by (4.169) that $$(b_{3}x_{7},\bar{b}_{3}b_{10})\neq 2$$. Therefore
$$(b_3x_7,\bar{b}_3b_{10})=1.$$
Now we can assume that
$$\bar{b}_3b_{10}=b_6+c+e\quad\hbox{where } e\in B,e\neq c\hbox{ and } e\neq d,$$
(4.171)
and by (4.10) we obtain that
$$b_3b_{15}=\bar{b}_{15}+b_6+c+e.$$
(4.172)
Now we obtain by (4.169) that $$(b_{15},\bar{b}_{3}c)=(x_{7},\bar{b}_{3}c)=1$$ which implies that
$$9\leq |c|\leq 17.$$
(4.173)
By (4.162), (4.169, (4.170) and $$b_{3}(\bar{b}_{3}x_{7})=\bar{b}_{3}(b_{3}x_{7})$$ we obtain that
$$\bar{b}_3c+\bar{b}_3d=b_{15}+x_7+x_8+b_3y_{11}.$$
(4.174)
Assume henceforth that (b 15,b 3 y 11)=1 and we shall derive a contradiction. Then we have that $$(b_{15},\bar{b}_{3}d)=1$$. By (4.171) we obtain that ed, by (4.169) and (4.3) we obtain that b 6d and by (4.173) and (4.169) we obtain that $$d\neq\bar{b}_{15}$$, which is a contradiction to (4.172). Now we have that
$$(b_{15},b_3y_{11})=0.$$
(4.175)
Assume henceforth that $$(\bar{b}_{15},\bar{b}_{3}y_{11})=1$$ and we shall derive a contradiction. Then by (4.12) we obtain that y 11 is a real element which implies that (b 15,b 3 y 11)=1, and we have a contradiction to (4.175). Now we have that
$$(\bar{b}_{15},\bar{b}_3y_{11})=0.$$
(4.176)
By (4.1), (4.169), (4.170), (4.171) and $$\bar{b}_{3}(\bar{b}_{3}x_{7})=\bar{b}_{3}^{2}x_{7}$$ we obtain that
$$b_6+e+\bar{b}_3y_{11}=d+\bar{b}_6x_7.$$
(4.177)
Assume that $$(u,\bar{b}_{3}c)=1$$ where uB. Then by (4.174) we obtain that |u|≥5.
Assume henceforth that u=u 5 and we shall derive a contradiction. Then by (4.174) we obtain that $$\bar{b}_{3}u_{5}=y_{11}+y_{4}$$ where y 4B and b 3 u 5=c+q where qB. Now by $$b_{3}(\bar{b}_{3}u_{5})=\bar{b}_{3}(b_{3}u_{5})$$ we obtain that $$\bar{b}_{3}c+\bar{b}_{3}q=b_{3}y_{11}+b_{3}y_{4}$$. Now by (4.172) we obtain that $$(b_{15},\bar{b}_{3}c)=1$$ and by (4.175) we obtain that (b 15,b 3 y 11)=0 which implies that (b 15,b 3 y 4)=1, and we have a contradiction. Hence
$$\hbox{If } (u,\bar{b}_3c)=1\quad\hbox{where } u\in B\quad\hbox{ then } |u|\geq 6.$$
(4.178)
Assume henceforth that c=c 9 and we shall derive a contradiction. Then by (4.169) and (4.172) we obtain that $$\bar{b}_{3}c_{9}=x_{7}+b_{15}+u_{5}$$ where u 5B, which is a contradiction to (4.178). Now (4.173) implies that
$$10\leq |c|\leq 17.$$
(4.179)
Assume henceforth that
$$(y_{11},b_3e)=1$$
(4.180)
and we shall derive a contradiction. Then by (4.171) and (4.177) we obtain that $$(b_{6}\bar{b}_{6},x_{7}\bar{x}_{7})=(\bar{b}_{6}x_{7},\bar{b}_{6}x_{7})\geq 6$$. By assumption $$(b_{3}\bar{b}_{3},x_{7}\bar{x}_{7})=2$$ which implies by (4.2) that $$(b_{8},x_{7}\bar{x}_{7})=1$$. Now by (4.164) we obtain that $$6\leq (b_{6}\bar{b}_{6},x_{7}\bar{x}_{7})=2+(s,x_{7}\bar{x}_{7})+(t,x_{7}\bar{x}_{7})$$. Hence $$(s,x_{7}\bar{x}_{7})+(t,x_{7}\bar{x}_{7})\geq 4$$ which implies that
$$\hbox{ either}\quad |s|\leq 10\quad\hbox{or}\quad |t|\leq 10.$$
(4.181)
By (4.3) and (4.169), we obtain that db 6 and by (4.171) we obtain that de. Now (4.177) implies that $$(d,\bar{b}_{3}y_{11})=1$$ and by (4.180) we obtain that
$$\bar{b}_3y_{11}=d+e+\varSigma \quad\hbox{where }\varSigma \in\mathbb{N}B.$$
(4.182)
Now (4.171) implies that
$$b_3e=b_{10}+y_{11}+C\quad\hbox{where } C\in\mathbb{N}B$$
(4.183)
and by (4.172) we obtain that
$$\bar{b}_3e=b_{15}+D\quad\hbox{where } D\in\mathbb{N}B.$$
(4.184)
Now by (4.171), (4.172) and $$b_{3}(\bar{b}_{3}e)=\bar{b}_{3}(b_{3}e)$$ we obtain that
$$\bar{b}_{15}+b_3D=\bar{b}_3y_{11}+\bar{b}_3C.$$
(4.185)
Now (4.176) implies that
$$(\bar{b}_{15},\bar{b}_3C)=(b_3\bar{b}_{15},C)=1.$$
(4.186)
By (4.165), (4.170), (4.183) and since b 10 is a non-real element we obtain that $$e\neq\bar{x}_{7}$$, $$e\neq\bar{x}_{8}$$ and $$e\neq\bar{b}_{15}$$. Thus $$(\bar{b}_{10},b_{3}e)=(\bar{b}_{3}\bar{b}_{10},e)=0$$ and $$(b_{8},b_{3}e)=(\bar{b}_{3}b_{8},e)=\nobreak 0$$. Now by (4.165), (4.183) and (4.186) we can assume that (s,C)=1 which implies that
$$(s,b_3e)=1\quad\hbox{and}\quad |s|\leq |C|.$$
(4.187)
By (4.183) we obtain that |C|=3|e|−21 which implies that
$$|s|\leq 3|e|-21.$$
(4.188)
By (4.165) and (4.187) we obtain that
$$(\bar{b}_3s,e)=(\bar{b}_3s,\bar{b}_{15})=1,$$
(4.189)
and together with (4.188) we obtain that 11≤|e|.
By (4.171) and (4.179) we obtain that |e|≤14. Therefore
$$11\leq |e|\leq 14.$$
(4.190)

Assume henceforth that e=e 11 and we shall derive a contradiction. Then by (4.189) we obtain that 11≤|s| and by (4.183) we obtain that |C|=12 which implies by ( 4.188 ) that |s|≤12. Hence 11≤|s|≤12 which implies by (4.164) that 15≤|t|≤16, and we have a contradiction to (4.181).

Assume henceforth that e=e 14 and we shall derive a contradiction. Then by (4.189) we obtain that 12≤|s| and by (4.183) we obtain that |C|=21 which implies by (4.188) that |s|≤21. Therefore
$$12\leq |s|\leq 21.$$
By (4.166) and (4.183) we obtain that 12≤|s|≤15, and by (4.164) we obtain that 12≤|t|≤15, which is a contradiction to (4.181).
Assume henceforth that e=e 12 and we shall derive a contradiction. Then by (4.189) we obtain that 9≤|s| and by (4.188) we obtain that |s|≤15. Hence 9≤|s|≤15. Now (4.164) and (4.181) imply that 9≤|s|≤10. By (4.189) it is impossible that s=s 10. Therefore s=s 9 and $$\bar{b}_{3}s_{9}=e_{12}+\bar{b}_{15}$$ and by (4.183) we obtain that
$$b_3e_{12}=s_9+y_{11}+b_{10}+w_6\quad\hbox{where } w_6\in B.$$
Now by (4.2), (4.165) and $$b_{3}(\bar{b}_{3}s_{9})=(b_{3}\bar{b}_{3})s_{9}$$ we obtain that
$$b_8s_9=b_8+s_9+b_{10}+w_6+\bar{b}_{10}+t_{18}+y_{11}.$$
Since b 8 and s 9 are real elements we obtain that y 11 is a real element. By (4.172) we obtain that c=c 12 and by (4.169) d=d 9. Now (4.170) and (4.182) imply that
$$\bar{b}_3y_{11}=e_{12}+d_9+\bar{x}_7+w_5\quad\hbox{where } w_5\in B.$$
(4.191)
Now we have that
$$b_3w_5=y_{11}+v_4\quad\hbox{where } v_4\in B$$
(4.192)
and by (4.174) we obtain that either $$(\bar{w}_{5},\bar{b}_{3}c_{12})=1$$ or $$(\bar{w}_{5},\bar{b}_{3}d_{9})=1$$. Now by (4.178) we obtain that $$(\bar{w}_{5},\bar{b}_{3}d_{9})=1$$ which implies that
$$\bar{b}_3w_5=\bar{d}_9+w_6\quad\hbox{where } w_6\in B.$$
Now by (4.191), (4.192) and $$\bar{b}_{3}(b_{3}w_{5})=b_{3}(\bar{b}_{3}w_{5})$$ we obtain that $$e_{12}+d_{9}+\bar{x}_{7}+w_{5}+\bar{b}_{3}v_{4}=b_{3}\bar{d}_{9}+b_{3}w_{6}$$. Since (w 5,b 3 w 6)=1 we obtain that (e 12,b 3 w 6)=0 which implies that $$(e_{12},b_{3}\bar{d}_{9})=1$$. By (4.169) we obtain that $$(\bar{x}_{7},b_{3}\bar{d}_{9})=1$$ and together with $$(w_{5},b_{3}\bar{d}_{9})=1$$, we obtain that
$$b_3\bar{d}_9=w_5+\bar{x}_7+e_{12}+r_3\quad\hbox{where } r_3\in B$$
which implies that $$(r_{3},\bar{b}_{3}v_{4})=1$$ and $$b_{3}\bar{r}_{3}=d_{9}$$, which is a contradiction.
Now we have that e=e 13 which implies by (4.188) that |s|≤18, and by (4.189) we obtain that 11≤|s|. Therefore 11≤|s|≤18. Now (4.181) and (4.164) imply that 17≤|s|≤18. By (4.183) and (4.187) we obtain that |s|≠17. Thus s=s 18 and by (4.165) we obtain that t=t 9 and
$$\bar{b}_3t_9=\bar{b}_{15}+E_{12}\quad\hbox{where }E_{12}\in\mathbb{N}B.$$
Now by (4.2), (4.165) and $$b_{3}(\bar{b}_{3}t_{9})=(b_{3}\bar{b}_{3})t_{9}$$ we obtain that
$$b_8t_9=b_8+\bar{b}_{10}+s_{18}+b_3E_{12}.$$
Now since b 8 and t 9 are real elements and b 10 is a non-real element, we obtain that $$(\bar{b}_{10},b_{3}E_{12})=1$$, which is a contradiction to (4.171).
Now we have that
$$(y_{11},b_3e)=0.$$
(4.193)
By (4.4), (4.162), (4.170), (4.171) and $$b_{3}(\bar{b}_{3}b_{10})=\bar{b}_{3}(b_{3}b_{10})$$ we obtain that
$$b_8+b_3c+b_3e=\bar{b}_3b_{15}+\bar{b}_3x_8+y_{11}.$$
(4.194)
Now (4.193) implies that (y 11,b 3 c)=1 and together with (4.171) we obtain that
$$b_3c=b_{10}+y_{11}+F\quad\hbox{where } F\in\mathbb{N}B.$$
(4.195)
Now by (4.177) we obtain that
$$\bar{b}_3y_{11}=c+d+G_{12}\quad\hbox{where }G_{12}\in\mathbb{N}B.$$
(4.196)
By (4.169) and (4.172) we obtain that
$$\bar{b}_3c=x_7+b_{15}+H\quad\hbox{where }H\in\mathbb{N}B.$$
(4.197)
Now by (4.169), (4.171), (4.172) and $$b_{3}(\bar{b}_{3}c)=\bar{b}_{3}(b_{3}c)$$ we obtain that
$$\bar{b}_{15}+b_3H=G_{12}+\bar{b}_3F.$$
(4.198)
If c=c 10 then by (4.197) we obtain that H=H 8B and (c 10,b 3 H 8)=1, which is a contradiction to (4.198), and if c=c 17 then (4.169) implies that d=d 4, which is a contradiction to (4.196). Now (4.179) implies that
$$11\leq |c|\leq 16.$$
From (4.169) and (4.171) we have seven cases (see Table 4.3). In the following seven sections, we derive a contradiction for each case in Table 4.3.
Table 4.3

Splitting to seven cases

 Case 1 c=c 11 d=d 10 e=e 13 Case 2 c=c 12 d=d 9 e=e 12 Case 3 c=c 13 d=d 8 e=e 11 Case 4 c=c 14 d=d 7 e=e 10 Case 5 c=c 15 d=d 6 e=e 9 Case 6 c=c 16 d=d 5 e=e 8 Case 7 (b 3 x 7,b 3 x 7)=3

### 4.5.1 Case c=c11

In this section we assume that c=c 11. Then (4.178) and (4.197) imply that
$$\bar{b}_3c_{11}=x_7+b_{15}+f_{11}\quad\hbox{where } f_{11}\in B.$$
(4.199)
Now (4.195) implies that
$$b_3c_{11}=b_{10}+y_{11}+g_{12}\quad\hbox{where }g_{12}\in B.$$
(4.200)
By (4.198) we obtain that
$$\bar{b}_{15}+b_3f_{11}=G_{12}+\bar{b}_3g_{12}$$
(4.201)
which implies that $$(g_{12},b_{3}\bar{b}_{15})=1$$, and by (4.165) we obtain that
$$b_3\bar{b}_{15}=b_8+\bar{b}_{10}+g_{12}+g_{15}\quad\hbox{where } g_{15}\in B.$$
(4.202)
By (4.170), (4.174) and (4.199) we obtain that
$$b_3y_{11}=x_7+f_{11}+\varSigma _{15}\quad\hbox{where}\varSigma _{15}\in\mathbb{N}B.$$
(4.203)
By (4.196) we obtain that
$$\bar{b}_3y_{11}=c_{11}+d_{10}+G_{12}.$$
(4.204)
Assume henceforth that y 11 is a real element and we shall derive a contradiction. Then by (4.203) and (4.204) we obtain that $$c_{11}=\bar{f}_{11}$$ and
$$b_3y_{11}=x_7+\bar{d}_{10}+\bar{c}_{11}+h_5\quad\hbox{where } h_5\in B$$
(4.205)
which implies that
$$\bar{b}_3h_5=y_{11}+h_4\quad\hbox{where }h_4\in B.$$
(4.206)
By (4.174), (4.199) and (4.205) we obtain that
$$\bar{b}_3d_{10}=x_7+x_8+\bar{d}_{10}+h_5$$
(4.207)
which implies that
$$b_3h_5=d_{10}+v_5\quad\hbox{where } v_5\in B.$$
(4.208)
Now $$b_{3}(\bar{b}_{3}h_{5})=\bar{b}_{3}(b_{3}h_{5})$$ implies that $$\bar{c}_{11}+b_{3}h_{4}=x_{8}+\bar{b}_{3}v_{5}$$. Hence $$(v_{5},\allowbreak b_{3}\bar{c}_{11})=\nobreak 1$$, which is a contradiction to (4.199).
Now we have that y 11 is a non-real element. By (4.200) and (4.201) we obtain that
$$\bar{b}_3g_{12}=\bar{b}_{15}+c_{11}+\varSigma _{10}\quad\hbox{where } \varSigma _{10}\in\mathbb{N}B.$$
(4.209)
Now by (4.2), (4.200), (4.202) and $$b_{3}(\bar{b}_{3}g_{12})=(b_{3}\bar{b}_{3})g_{12}$$, we obtain that
$$b_8g_{12}=b_8+b_{10}+\bar{b}_{10}+y_{11}+g_{12}+g_{15}+b_3\varSigma _{10}.$$
(4.210)
Since b 8 and g 12 are real elements and y 11 is a non-real element, we obtain that $$(\bar{y}_{11},b_{3}\varSigma _{10})=1$$. Now (4.209) implies that Σ 10B and by (4.203) we obtain that
$$b_3y_{11}=\bar{ \varSigma }_{10}+x_7+f_{11}+r_5\quad\hbox{where } r_5\in B.$$
(4.211)
Now by (4.174) and (4.199) we obtain that
$$\bar{b}_3d_{10}=r_5+x_7+x_8+\bar{ \varSigma }_{10}.$$
(4.212)
Now
$$\bar{b}_3r_5=y_{11}+h_4\quad\hbox{where } h_4\in B$$
and
$$b_3r_5=d_{10}+v_5\quad\hbox{where } v_5\in B.$$
(4.213)
By $$b_{3}(\bar{b}_{3}r_{5})=\bar{b}_{3}(b_{3}r_{5})$$ we obtain that
$$f_{11}+b_3h_4=x_8+\bar{b}_3v_5$$
which implies that $$(f_{11},\bar{b}_{3}v_{5})=1$$, and we have a contradiction to (4.213).

### 4.5.2 Case c=c12

In this section we assume that c=c 12. Then (4.195) and (4.197) imply that
$$\bar{b}_3c_{12}=x_7+b_{15}+H_{14}$$
(4.214)
and
$$b_3c_{12}=b_{10}+y_{11}+F_{15}.$$
(4.215)
Assume henceforth that y 11 is a real element and we shall derive a contradiction. Then by (4.170) and (4.196) we obtain that
$$b_3y_{11}=x_7+\bar{d}_9+\bar{c}_{12}+h_5\quad\hbox{where } h_5\in B.$$
(4.216)
Now by (4.174) and (4.214) we obtain that
$$H_{14}+\bar{b}_3d_9=x_8+\bar{c}_{12}+\bar{d}_9+x_7+h_5$$
which implies that
$$\bar{b}_3c_{12}=x_7+b_{15}+\bar{d}_9+h_5,$$
and we have a contradiction to (4.178).
Now we have that y 11 is a non-real element. By (4.198) we obtain that
$$\bar{b}_{15}+b_3H_{14}=G_{12}+\bar{b}_3F_{15},$$
(4.217)
which implies that $$\bar{b}_{15}\in \mathit{Irr}(\bar{b}_{3}F_{15})$$. By (4.5) we obtain that (b 8,F 15)=(b 8,b 3 c 12)=0, and by (4.162) we obtain that $$(\bar{b}_{10},F_{15})=(\bar{b}_{10},b_{3}c_{12})=0$$. Hence by (4.165) we can assume that
$$s\in \mathit{Irr}(F_{15})$$
(4.218)
which implies that |s|≤15, and by (4.164) s is a real element. Now we have that
$$(s,b_3c_{12})=1$$
(4.219)
and by (4.165) we obtain that
$$(s,b_3\bar{b}_{15})=1$$
(4.220)
which implies that 9≤|s|. Hence
$$9\leq |s|\leq 15.$$
(4.221)
Assume henceforth that s=s 15 and we shall derive a contradiction. Then by (4.215) and (4.218) we obtain that
$$b_3c_{12}=b_{10}+y_{11}+s_{15}.$$
(4.222)
Now (4.165) implies that
$$\bar{b}_3s_{15}=c_{12}+\bar{b}_{15}+\varSigma _{18}\quad\hbox{where } \varSigma _{18}\in\mathbb{N}B$$
(4.223)
and (4.214) implies that
$$\bar{b}_3c_{12}=x_7+b_{15}+f_{14}\quad\hbox{where } f_{14}\in B.$$
(4.224)
Now by (4.2), (4.165) and $$b_{3}(\bar{b}_{3}s_{15})=(b_{3}\bar{b}_{3})s_{15}$$ we obtain that
$$b_8s_{15}=b_8+b_{10}+\bar{b}_{10}+y_{11}+t_{12}+s_{15}+b_3\varSigma _{18}.$$
Since b 8 and s 15 are real elements and y 11 is a non-real element, we obtain that
$$\bar{y}_{11}\in \mathit{Irr}(b_3\varSigma _{18}).$$
By (4.174) we obtain that (f 14,b 3 y 11)=1, and together with (4.170) we obtain that
$$b_3y_{11}=x_7+f_{14}+S_{12}\quad\hbox{where }S_{12}\in\mathbb{N}B.$$
(4.225)
Since $$\bar{y}_{11}\in \mathit{Irr}(b_{3}\varSigma _{18})$$ we obtain that there exists an element α in B such that $$\alpha\in \mathit{Irr}(\bar{ \varSigma }_{18})\cap \mathit{Irr}(S_{12})$$ which implies that 10≤|α|. Hence S 12=α 12. Now we have that
$$\bar{b}_3s_{15}=c_{12}+\bar{b}_{15}+\bar{\alpha}_{12}+\alpha_{6}\quad\hbox{where }\alpha_6\in B$$
(4.226)
which implies that
$$b_3\alpha_6=s_{15}+\alpha_3\quad\hbox{where } \alpha_3\in B,$$
(4.227)
and by (4.225) we obtain that
$$b_3y_{11}=x_7+f_{14}+\alpha_{12}.$$
By (4.196) we obtain that
$$\bar{b}_3y_{11}=c_{12}+d_9+\beta_{12}\quad\hbox{where } \beta_{12}\in B.$$
By (4.169), (4.171) (4.172), (4.224), (4.226) and $$b_{3}(\bar{b}_{3}c_{12})=\bar{b}_{3}(b_{3}c_{12})$$, we obtain that
$$b_3f_{14}=\beta_{12}+c_{12}+\bar{\alpha}_{12}+\alpha_6.$$
Now we have that
$$\bar{b}_3\alpha_6=f_{14}+\alpha_4\quad\hbox{where }\alpha_4\in B.$$
Now by (4.226), (4.227) and $$b_{3}(\bar{b}_{3}\alpha_{6})=\bar{b}_{3}(b_{3}\alpha_{6})$$ we obtain that $$\bar{b}_{15}+\bar{b}_{3}\alpha_{3}=\beta_{12}+b_{3}\alpha_{4}$$, and we have a contradiction.
Now we have that |s|≠15, so (4.215), (4.219) and (4.221) imply that
$$b_3c_{12}=b_{10}+y_{11}+s+k\quad\hbox{where } k\in B\hbox{ and }9\leq |s|\leq 11.$$
Now (4.219) and (4.220) imply that
$$\bar{b}_3s=c_{12}+\bar{b}_{15}+T\quad\hbox{where } T\in\mathbb{N}B\hbox { and } |T|\leq 6,$$
(4.228)
and together with (4.165) and $$b_{3}(\bar{b}_{3}s)=(b_{3}\bar{b}_{3})s$$ we obtain that
$$b_8s=b_8+b_{10}+\bar{b}_{10}+y_{11}+s+t+k+b_3T.$$
Since b 8 and s are real elements and y 11 is a non-real element, we obtain that $$\bar{y}_{11}\in \mathit{Irr}(b_{3}T)$$, which is a contradiction to (4.228).

### 4.5.3 Case c=c13

In this section we assume that c=c 13. Then (4.195) and (4.197) imply that
$$\bar{b}_3c_{13}=x_7+b_{15}+H_{17}$$
(4.229)
and
$$b_3c_{13}=b_{10}+y_{11}+F_{18}.$$
(4.230)
Assume henceforth that y 11 is a real element and we shall derive a contradiction. Then by (4.170) and (4.196) we obtain that
$$b_3y_{11}=x_7+\bar{d}_8+\bar{c}_{13}+h_5\quad\hbox{where } h_5\in B.$$
(4.231)
Now by (4.174) and (4.229) we obtain that
$$H_{17}+\bar{b}_3d_8=x_8+\bar{c}_{13}+\bar{d}_8+x_7+h_5,$$
and we have a contradiction.
Now we have that y 11 is a non-real element. By (4.198) we obtain that
$$\bar{b}_{15}+b_3H_{17}=G_{12}+\bar{b}_3F_{18}.$$
Now we have that $$\bar{b}_{15}\in \mathit{Irr}(\bar{b}_{3}F_{18})$$. By (4.5), (4.162) and (4.230) we can assume that
$$s\in \mathit{Irr}(F_{18})$$
(4.232)
and together with (4.165) we obtain that
$$11\leq |s|\leq 18.$$
(4.233)
Assume henceforth that s=s 18 and we shall derive a contradiction. Then (4.230) and (4.232) imply that
$$b_3c_{13}=b_{10}+y_{11}+s_{18}.$$
(4.234)
By (4.229) we obtain that
$$\bar{b}_3c_{13}=x_7+b_{15}+f_{17}$$
(4.235)
and by (4.165) we obtain that
$$b_3\bar{b}_{15}=b_8+\bar{b}_{10}+s_{18}+t_9.$$
(4.236)
Now we have that
$$\bar{b}_3s_{18}=c_{13}+\bar{b}_{15}+\varSigma _{26}\quad\hbox{where } \varSigma _{26}\in\mathbb{N}B.$$
(4.237)
By (4.2) and $$b_{3}(\bar{b}_{3}s_{18})=(b_{3}\bar{b}_{3})s_{18}$$ we obtain that
$$b_8s_{18}=b_8+t_9+b_{10}+\bar{b}_{10}+y_{11}+s_{18}+b_3\varSigma _{26}.$$
Since b 8 and s 18 are real elements and y 11 is a non-real element, we obtain that
$$\bar{y}_{11}\in \mathit{Irr}(b_3\varSigma _{26}).$$
(4.238)
By (4.174) and (4.235) we obtain that
$$\bar{b}_3d_8=x_7+x_8+g_9\quad\hbox{where }g_9\in B$$
(4.239)
and
$$b_3y_{11}=x_7+f_{17}+g_9.$$
(4.240)
Now (4.196) implies that
$$\bar{b}_3y_{11}=c_{13}+d_8+h_{12}\quad\hbox{where } h_{12}\in B.$$
(4.241)
By (4.237) and (4.240) we obtain that $$\bar{g}_{9}\notin \varSigma _{26}$$, and by (4.170) and (4.237) we obtain that $$\bar{x}_{7}\notin \varSigma _{26}$$. Now (4.238) and (4.240) imply that $$\bar{f}_{17}\in \varSigma _{26}$$. By (4.237) we obtain that
$$\bar{b}_3s_{18}=c_{13}+\bar{b}_{15}+\bar{f}_{17}+f_9\quad\hbox{where } f_9\in B.$$
(4.242)
By (4.198) we obtain that
$$b_3f_{17}=h_{12}+c_{13}+\bar{f}_{17}+f_9$$
(4.243)
which implies that
$$\bar{b}_3f_9=f_{17}+\varSigma _{10}\quad\hbox{where } \varSigma _{10}\in\mathbb{N}B$$
(4.244)
and
$$b_3f_9=s_{18}+\varSigma _9\quad\hbox{where }\varSigma _9\in\mathbb{N}B.$$
Now by (4.242), (4.243) and $$b_{3}(\bar{b}_{3}f_{9})=\bar{b}_{3}(b_{3}f_{9})$$ we obtain that
$$h_{12}+b_3\varSigma _{10}=\bar{b}_{15}+\bar{b}_3\varSigma _9$$
which implies that $$\bar{b}_{15}\in \mathit{Irr}(b_{3}\varSigma _{10})$$, and by (4.172) we obtain that $$\bar{b}_{6}\in \varSigma _{10}$$. Now (4.244) implies that $$(f_{9},b_{3}\bar{b}_{6})=1$$, which is a contradiction to (4.3).
Now we have that |s|≠18 which implies by (4.230) and (4.233) that 11≤|s|≤12. By (4.165), (4.230) and (4.232) we obtain that
$$\bar{b}_3s=c_{13}+\bar{b}_{15}+T\quad\hbox{where } T\in\mathbb{N}B\hbox{ and } |T|\leq 8.$$
(4.245)
Now by (4.2), (4.230), (4.165) and $$b_{3}(\bar{b}_{3}s)=(b_{3}\bar{b}_{3})s$$ we obtain that
$$b_8s=b_8+b_{10}+\bar{b}_{10}+y_{11}+t+F_{18}+b_3T.$$
Since b 8 and s are real elements and y 11 is a non-real element, we obtain that $$s\neq\bar{y}_{11}$$; and by (4.232) and 11≤|s|≤12, we have that $$\bar{y}_{11}\notin F_{18}$$. Hence $$\bar{y}_{11}\in b_{3}T$$, which is a contradiction to (4.245).

### 4.5.4 Case c=c14

In this section we assume that c=c 14. Then (4.195) and (4.197) imply that
$$\bar{b}_3c_{14}=x_7+b_{15}+H_{20}$$
(4.246)
and
$$b_3c_{14}=b_{10}+y_{11}+F_{21}.$$
(4.247)
By (4.169) we obtain that
$$\bar{b}_3d_7=x_7+\varSigma _{14}\quad\hbox{where }\varSigma _{14}\in\mathbb{N}B.$$
(4.248)
By (4.198) we obtain that
$$\bar{b}_{15}+b_3H_{20}=G_{12}+\bar{b}_3F_{21}$$
which implies that $$\bar{b}_{15}\in \mathit{Irr}(\bar{b}_{3}F_{21})$$. By (4.5) and (4.162) we obtain that b 8Irr(b 3 c 14) and $$\bar{b}_{10}\notin \mathit{Irr}(b_{3}c_{14})$$, and by (4.165) and (4.247) we can assume that
$$s\in \mathit{Irr}(F_{21}).$$
(4.249)
By (4.165) we obtain that $$(s,b_{3}\bar{b}_{15})=1$$ and together with (4.166) we obtain that
$$12\leq |s|\leq 15$$
(4.250)
and
$$\bar{b}_3s=\bar{b}_{15}+c_{14}+T\quad\hbox{where } T\in\mathbb{N}B.$$
(4.251)
Assume henceforth that y 11 is a real element and we shall derive a contradiction. Then by (4.170) and (4.196) we obtain that either
$$b_3y_{11}=x_7+\bar{d}_7+\bar{c}_{14}+h_5\quad\hbox{where } h_5\in B$$
(4.252)
or $$x_{7}=\bar{d}_{7}$$ and then
$$b_3y_{11}=x_7+\bar{c}_{14}+\varSigma _{12}\quad\hbox{where }\varSigma _{12}\in\mathbb{N}B.$$
(4.253)
If (4.252) holds then by (4.174) and (4.246), we obtain that
$$H_{20}+\bar{b}_3d_7=x_8+\bar{c}_{14}+\bar{d}_7+x_7+h_5.$$
By (4.169) we obtain that $$(x_{7},\bar{b}_{3}d_{7})=1$$ which implies that $$H_{20}=x_{8}+\bar{d}_{7}+h_{5}$$, and we have a contradiction to (4.246).
Now we have that $$x_{7}=\bar{d}_{7}$$. By (4.247), (4.249) and (4.250) we obtain that
$$b_3c_{14}=b_{10}+y_{11}+s+g\quad\hbox{where }g\in B\hbox{ and } 6\leq |g|\leq 9.$$
(4.254)
By (4.162), (4.169), (4.170), (4.246), (4.252) and $$b_{3}(\bar{b}_{3}x_{7})=\bar{b}_{3}(b_{3}x_{7})$$, we obtain that
$$x_8+\varSigma _{12}=H_{20}.$$
By (4.254) we obtain that (b 3 c 14,b 3 c 14)=4. Therefore Σ 12=h 12 where h 12B and H 20=x 8+h 12. Therefore
$$\bar{b}_3c_{14}=x_7+b_{15}+x_8+h_{12}$$
(4.255)
and
$$b_3y_{11}=x_7+\bar{c}_{14}+h_{12}.$$
(4.256)
By (4.194), (4.165) and (4.254) we obtain that
$$g+b_3e_{10}=t+\bar{b}_3x_8.$$
It is obvious that gt. By (4.165) and (4.250) we obtain that 12≤|t|≤15. Now (4.171) implies that
$$b_3e_{10}=b_{10}+t+k\quad\hbox{where } k\in B$$
(4.257)
and
$$\bar{b}_3x_8=b_{10}+g+k.$$
(4.258)
By (4.255) we obtain that
$$b_3x_8=c_{14}+\alpha+\beta\quad\hbox{where}\alpha,\beta\in B.$$
(4.259)
Assume henceforth that g is a real element and we shall derive a contradiction. By (4.254) and (4.258) we obtain that $$(\bar{c}_{14},b_{3}g)=(x_{8},b_{3}g)$$ which implies that 9≤|g|. By (4.254) we obtain that g=g 9. Now we have that
$$b_3g_9=x_8+\bar{c}_{14}+w_5\quad\hbox{where } w_5\in B.$$
By (4.162), (4.255), (4.258), (4.258), (4.259) and $$b_{3}(\bar{b}_{3}x_{8})=\bar{b}_{3}(b_{3}x_{8})$$, we obtain that $$x_{8}+\bar{c}_{14}+w_{5}+b_{3}k_{5}=h_{12}+\bar{b}_{3}\alpha+\bar{b}_{3}\beta$$. Hence (h 12,b 3 k 5)=1, which is a contradiction to (4.258).
Now we have that g is a non-real element. By (4.2), (4.256), (4.170), (4.254) and $$(b_{3}\bar{b}_{3})y_{11}=b_{3}(\bar{b}_{3}y_{11})$$, we obtain that
$$b_8y_{11}=b_{10}+\bar{b}_{10}+y_{11}+s+g+b_3\bar{h}_{12}.$$
Since b 8 and y 11 are real elements and g is a non-real element, we obtain that $$(h_{12},b_{3}g)=(\bar{g},b_{3}\bar{h}_{12})=1$$ and by (4.258) we obtain that (x 8,b 3 g)=1 which implies that 8≤|g|. By (4.254) we obtain that either g=g 8 or g=g 9.
Assume henceforth that g=g 9 and we shall derive a contradiction. Then by (4.258) and (4.257) we obtain that
$$b_3k_5=x_8+h_7\quad\hbox{where } h_7\in B$$
and
$$\bar{b}_3k_5=e_{10}+h_5\quad\hbox{where }h_5\in B.$$
(4.260)
Now by (4.258), (4.257) and $$b_{3}(\bar{b}_{3}k_{5})=\bar{b}_{3}(b_{3}k_{5})$$, we obtain that $$t_{15}+b_{3}h_{5}=g_{9}+\bar{b}_{3}h_{7}$$. Hence (g 9,b 3 h 5)=1, which is a contradiction to (4.260).
Now we have that g=g 8. By (4.258) we obtain that (x 8,b 3 g 8)=1, and since (h 12,b 3 g 8)=1 we obtain that
$$b_3g_8=x_8+h_{12}+h_4\quad\hbox{where } h_4\in B.$$
By (4.254) we obtain that
$$\bar{b}_3g_8=c_{14}+m+n\quad\hbox{where } m,n\in B.$$
Now by (4.254), (4.258) and $$b_{3}(\bar{b}_{3}g_{8})=\bar{b}_{3}(b_{3}g_{8})$$, we obtain that $$y_{11}+s_{13}+b_{3}m+b_{3}n=k_{6}+\bar{b}_{3}h_{4}+\bar{b}_{3}h_{12}$$ which implies that $$\bar{b}_{3}h_{12}=g_{8}+y_{11}+s_{13}+\gamma_{4}$$ where γ 4B. Hence b 3 γ 4=h 12 which implies that (b 3 γ 4,b 3 γ 4)=1. On the other hand, we have that either $$(m,\bar{b}_{3}\gamma_{4})=1$$ or $$(n,\bar{b}_{3}\gamma_{4})=1$$ which implies that (b 3 γ 4,b 4 γ 4)≠1, and we have a contradiction.
Now we have that y 11 is a non-real element. By (4.2), (4.165), (4.247), (4.251) and $$b_{3}(\bar{b}_{3}s)=(b_{3}\bar{b}_{3})s$$ we obtain that
$$b_8s=b_8+b_{10}+\bar{b}_{10}+s+t+F_{21}+b_3T.$$
By (4.249) and (4.250) we obtain that $$\bar{y}_{11}\notin \mathit{Irr}(F_{21})$$. Now since b 8 and s are real elements and y 11 is a non-real element we obtain that
$$\bar{y}_{11}\in \mathit{Irr}(b_3T).$$
(4.261)
If s=s 12 then by (4.251) we obtain that
$$\bar{b}_3s_{12}=\bar{b}_{15}+c_{14}+f_7\quad\hbox{where } f_7\in B.$$
Now we have that $$(\bar{y}_{11},b_{3}f_{7})=(s_{12},b_{3}f_{7})=1$$, which is a contradiction.
If s=s 13 then by (4.251), we obtain that
$$\bar{b}_3s_{13}=\bar{b}_{15}+c_{14}+f_{10}\quad\hbox{where } f_{10}\in B.$$
(4.262)
Now (4.261) implies that
$$b_3f_{10}=s_{13}+\bar{y}_{11}+h_6\quad\hbox{where } h_6\in B.$$
(4.263)
By (4.170) we obtain that
$$b_3y_{11}=x_7+\bar{f}_{10}+\varSigma _{16}\quad\hbox{where }\varSigma _{16}\in\mathbb{N}B.$$
Now by (4.174), (4.248) and (4.246), we obtain that $$H_{20}+\varSigma _{14}=x_{8}+\bar{f}_{10}+\varSigma _{16}$$ which implies that
$$\begin{array}{*{20}l} {\bar b_3 c_{14}= x_7+ \bar f_{10}+ g_{10}+ b_{15} } & {{\rm where}\,g_{10}\in B,}\\ \end{array}$$
(4.264)
$$\begin{array}{*{20}l} {b_3 y_{11}= x_7+ \bar f_{10}+ g_{10}+ g_6 } & {{\rm where}\,g_6\in B}\\ \end{array}$$
(4.265)
and
$$\bar{b}_3d_7=x_7+x_8+g_6.$$
(4.266)
By (4.247) and (4.249) we obtain that
$$b_3c_{14}=b_{10}+y_{11}+s_{13}+g_8\quad\hbox{where } g_8\in B.$$
By (4.198) and (4.262) we obtain that
$$b_3\bar{f}_{10}+b_3g_{10}=G_{12}+c_{14}+f_{10}+\bar{b}_3g_8.$$
By (4.264) we obtain that
$$\bar{b}_3f_{10}=\bar{c}_{14}+Q_{16}\quad\hbox{where } Q_{16}\in\mathbb{N}B.$$
(4.267)
Now by (4.264), (4.263), (4.262), (4.265) and $$b_{3}(\bar{b}_{3}f_{10})=\bar{b}_{3}(b_{3}f_{10})$$ we obtain that
$$c_{14}+f_{10}+\bar{g}_6+\bar{b}_3h_6=b_3Q_{16}$$
(4.268)
which implies that c 14Irr(b 3 Q 16).
By (4.264) one of the following cases hold:
1. Case 1:

x 7Irr(Q 16).

2. Case 2:

g 10Irr(Q 16).

3. Case 3:

$$\bar{f}_{10}\in \mathit{Irr}(Q_{16})$$.

Case 1 By (4.169) and (4.268) we obtain that $$(d_{7},\bar{b}_{3}h_{6})=1$$ and by (4.263) we obtain that $$(f_{10},\bar{b}_{3}h_{6})=1$$, which is a contradiction.

Case 2 By (4.267) we obtain that
$$\bar{b}_3f_{10}=\bar{c}_{14}+g_{10}+\alpha_6\quad\hbox{where}\alpha_6\in B.$$
Now (4.264) implies that
$$b_3g_{10}=\beta_6+f_{10}+c_{14}\quad\hbox{where}\beta_6\in B.$$
(4.269)
By (4.263) we obtain that $$(f_{10},\bar{b}_{3}h_{6})=1$$ which implies that $$(\beta_{6},\bar{b}_{3}h_{6})=0$$. Since g 10Irr(Q 16) we obtain by (4.268) and (4.269) that $$\beta_{6}=\bar{g}_{6}$$. Now by (4.269) we obtain that $$(\bar{g}_{10},b_{3}g_{6})=1$$, and by (4.266) we obtain that (d 7,b 3 g 6)=1, which is a contradiction.
Case 3 By (4.267) we obtain that
$$\bar{b}_3f_{10}=\bar{c}_{14}+\bar{f}_{10}+\gamma_6\quad\hbox{where}\gamma_6\in B.$$
By (4.263) we obtain that $$(f_{10},\bar{b}_{3}h_{6})=1$$ which implies that $$(\gamma_{6},\bar{b}_{3}h_{6})=1$$. Since $$\bar{f}_{10}\in Q_{16}$$ we obtain by (4.268) that $$\gamma_{6}=\bar{g}_{6}$$ which implies that $$(f_{10},b_{3}\bar{g}_{6})=1$$, and by (4.265) we obtain that $$(\bar{y}_{11},b_{3}\bar{g}_{6})=1$$, which is a contradiction.
If s=s 14 then by (4.251) we obtain that
$$\bar{b}_3s_{14}=c_{14}+\bar{b}_{15}+T_{13}.$$
Now by (4.261) we obtain that there exists an element fB in Irr(T 13) such that 10≤|f| which implies that f=T 13. Hence
$$\bar{b}_3s_{14}=c_{14}+\bar{b}_{15}+f_{13}\quad\hbox{where } f_{13}\in B.$$
(4.270)
By (4.170) and (4.261) we obtain that
$$b_3y_{11}=x_7+\bar{f}_{13}+\varSigma _{13}\quad\hbox{where }\varSigma _{13}\in\mathbb{N}B.$$
(4.271)
Now by (4.174), (4.246) and (4.248) we obtain that $$H_{20}+\varSigma _{14}=x_{8}+\bar{f}_{13}+\varSigma _{13}$$ which implies that
$$\begin{array}{c}\bar{b}_3c_{14}=x_7+g_7+\bar{f}_{13}+b_{15}\quad\hbox{where } g_7\in B,\\[6pt]\bar{b}_3d_7=x_7+x_8+g_6\quad\hbox{where } g_6\in B\end{array}$$
(4.272)
and
$$b_3y_{11}=\bar{f}_{13}+x_7+g_6+g_7.$$
(4.273)
By (4.247) and (4.249) we obtain that
$$b_3c_{14}=b_{10}+y_{11}+s_{14}+h_7\quad\hbox{where } h_7\in B.$$
Now we have that
$$b_3g_7=c_{14}+\alpha_7\quad\hbox{where }\alpha_7\in B$$
and
$$\bar{b}_3h_7=c_{14}+\beta_7\quad\hbox{where }\beta_7\in B.$$
By (4.198) we obtain that
$$b_3\bar{f}_{13}+\alpha_7=G_{12}+c_{14}+f_{13}+\beta_7.$$
(4.274)
If α 7Irr(G 12) then there exists an element α 5B such that α 5Irr(G 12) which implies that $$(\alpha_{5},b_{3}\bar{f}_{13})=1$$, and we have a contradiction. Hence α 7=β 7. By (4.273) we obtain that 3≤(b 3 y 11,b 3 y 11). Now (4.196) and (4.274) imply that
$$\bar{b}_3y_{11}=c_{14}+d_7+\alpha_6+\beta_6\quad\hbox{where}\alpha_6,\beta_6\in B$$
(4.275)
and
$$b_3\bar{f}_{13}=c_{14}+f_{13}+\alpha_6+\beta_6.$$
(4.276)
By (4.270) and (4.273) we obtain that
$$b_3f_{13}=\bar{y}_{11}+s_{14}+\varSigma _{14}\quad\hbox{where}\varSigma _{14}\in\mathbb{N}B.$$
Now by (4.270), (4.272), (4.273) and $$b_{3}(\bar{b}_{3}f_{13})=\bar{b}_{3}(b_{3}f_{13})$$ we obtain that
$$\bar{g}_6+\bar{b}_3\varSigma _{14}=\alpha_6+\beta_6+b_3\bar{\alpha}_6+b_3\bar{\beta}_6.$$
By (4.276) we obtain that $$(f_{13},b_{3}\bar{\alpha}_{6})=(f_{13},b_{3}\bar{\beta}_{6})=1$$ which implies that $$(\bar{g}_{6},b_{3}\bar{\alpha}_{6})=(\bar{g}_{6},b_{3}\bar{\beta}_{6})=0$$. Now we have that either $$g_{6}=\bar{\alpha}_{6}$$ or $$g_{6}=\bar{\beta}_{6}$$. By (4.273) we obtain that $$(\bar{y}_{11},b_{3}\bar{g}_{6})=1$$, and by (4.275) we obtain that $$(y_{11},\allowbreak b_{3}\bar{g}_{6})=\nobreak 1$$, which is a contradiction.
If s=s 15 then by (4.251) we obtain that
$$\bar{b}_3s_{15}=\bar{b}_{15}+c_{14}+T_{16}.$$
Now by (4.261) we obtain that there exists an element fB in Irr(T 16) such that 10≤|f| which implies that f=T 16. Hence
$$\bar{b}_3s_{15}=c_{14}+\bar{b}_{15}+f_{16}\quad\hbox{where } f_{16}\in B.$$
(4.277)
By (4.261) and (4.170) we obtain that
$$b_3y_{11}=x_7+\bar{f}_{16}+\varSigma _{10}\quad\hbox{where }\varSigma _{10}\in\mathbb{N}B.$$
(4.278)
Now by (4.174), (4.246) and (4.248) we obtain that $$H_{20}+\varSigma _{14}=x_{8}+\bar{f}_{16}+\varSigma _{10}$$ which implies that $$\bar{f}_{16}\in \mathit{Irr}(H_{20})$$ which implies by (4.246) that $$\bar{b}_{3}c_{14}=x_{7}+b_{15}+\bar{f}_{16}+f_{4}$$ where f 4B, and we have a contradiction since $$(f_{4},\bar{b}_{3}c_{14})=1$$.

### 4.5.5 Case c=c15

In this section we assume that c=c 15. Then (4.195) and (4.197) imply that
$$\bar{b}_3c_{15}=x_7+b_{15}+H_{23}$$
(4.279)
and
$$b_3c_{15}=b_{10}+y_{11}+F_{24}.$$
(4.280)
By (4.169) we obtain that
$$\bar{b}_3d_6=x_7+\varSigma _{11}\quad\hbox{where }\varSigma _{11}\in\mathbb{N}B.$$
(4.281)
Assume henceforth that y 11 is a real element and we shall derive a contradiction. Then by (4.170) and (4.196) we obtain that
$$b_3y_{11}=x_7+\bar{d}_6+\bar{c}_{15}+h_5\quad\hbox{where } h_5\in B.$$
(4.282)
Now by (4.174) and (4.279) we obtain that
$$H_{23}+\varSigma _{11}=x_8+\bar{c}_{15}+\bar{d}_6+h_5.$$
Now we have that $$H_{23}=x_{8}+\bar{c}_{15}$$ and $$\varSigma _{11}=\bar{d}_{6}+h_{5}$$ which imply by (4.279) and (4.281) that
$$\bar{b}_3c_{15}=x_7+x_8+b_{15}+\bar{c}_{15}$$
and
$$\bar{b}_3d_6=x_7+\bar{d}_6+h_5.$$
(4.283)
Now we have by (4.198) that $$\bar{b}_{15}+b_{3}x_{8}+\bar{x}_{7}+\bar{x}_{8}+\bar{b}_{15}+c_{15}=\bar{x}_{7}+\bar{h}_{5}+\bar{b}_{3}F_{24}$$ which implies that $$(\bar{x}_{8},b_{3}h_{5})=(\bar{h}_{5},b_{3}x_{8})=1$$, and by (4.283) we obtain that (d 6,b 3 h 5)=1, which is a contradiction.
Now we have that y 11 is a non-real element. By (4.198) we obtain that
$$\bar{b}_{15}+b_3H_{23}=G_{12}+\bar{b}_3F_{24}$$
which implies that $$\bar{b}_{15}\in \mathit{Irr}(\bar{b}_{3}F_{24})$$. By (4.280) we obtain that (y 11,b 3 c 15)=1, and by (4.165) and since y 11 is a non-real element we obtain that $$(y_{11},b_{3}\bar{b}_{15})=0$$ which implies that $$c_{15}\neq\bar{b}_{15}$$. Now by (4.5) and (4.162) we obtain that b 8Irr(b 3 c 15) and $$\bar{b}_{10}\notin \mathit{Irr}(b_{3}c_{15})$$ and by (4.165) (4.280) we can assume that
$$s\in \mathit{Irr}(F_{24}).$$
(4.284)
By (4.165) we obtain that $$(s,b_{3}\bar{b}_{15})=1$$ which implies that
$$\bar{b}_3s=\bar{b}_{15}+c_{15}+T\quad\hbox{where } T\in\mathbb{N}B$$
(4.285)
which implies that
$$10\leq |s|;$$
(4.286)
and by (4.280) we obtain that
$$b_3c_{15}=b_{10}+y_{11}+s+S\quad\hbox{where }S\in\mathbb{N}B.$$
(4.287)
Now by (4.2), (4.165) and $$b_{3}(\bar{b}_{3}s)=(b_{3}\bar{b}_{3})s$$ we obtain that
$$b_8s=b_8+b_{10}+\bar{b}_{10}+y_{11}+s+t+S+b_3T.$$
Since b 8 and s are real elements and y 11 is a non-real element we obtain that either $$\bar{y}_{11}\in \mathit{Irr}(S)$$ or $$\bar{y}_{11}\in \mathit{Irr}(b_{3}T)$$. By (4.174) we obtain that
$$\bar{b}_3c_{15}+\bar{b}_3d_6=b_{15}+x_7+x_8+b_3y_{11}.$$
If $$(x_{8},\bar{b}_{3}d_{6})=1$$ then by ( 4.169 ), we obtain that
$$\bar{b}_3d_6=x_7+x_8+\alpha_3\quad\hbox{where } \alpha_3\in B$$
which implies that (α 3,b 3 y 11)=1, and we have a contradiction. Hence $$(x_{8},\allowbreak\bar{b}_{3}c_{15})=\nobreak 1$$ which implies that
$$b_3x_8=c_{15}+\varSigma _9\quad\hbox{where }\varSigma _9\in\mathbb{N}B.$$
(4.288)
In addition, we have that
$$2\leq (b_3x_8,b_3x_8)\leq 3.$$
If $$\bar{y}_{11}\in S$$ then we obtain by (4.286) and by (4.287) that
$$b_3c_{15}=b_{10}+y_{11}+\bar{y}_{11}+s_{13}.$$
Now by (4.285) we obtain that
$$\bar{b}_3s_{13}=\bar{b}_{15}+c_{15}+\alpha_9\quad\hbox{where }\alpha_9\in B.$$
By (4.170) and since $$\bar{y}_{11}\in S$$ we obtain that
$$b_3y_{11}=x_7+\bar{c}_{15}+\varSigma ^*_{11}.$$
By (4.174), (4.279), (4.281) and (4.288) we obtain that
$$\bar{b}_3c_{15}=x_7+x_8+b_{15}+\bar{c}_{15}.$$
Now by (4.169), (4.171), (4.172), (4.196), (4.288) and $$b_{3}(\bar{b}_{3}c_{15})=\bar{b}_{3}(b_{3}c_{15})$$ implies that $$\varSigma _{9}+\bar{x}_{8}+\bar{b}_{15}=\varSigma _{12}+\alpha_{9}+\bar{ \varSigma }^{*}_{11}$$, which is a contradiction.

Now we have that $$\bar{y}_{11}\in b_{3}T$$ which implies that 5≤|T| and by (4.285) we obtain that 12≤|s|.

Assume henceforth that (b 3 x 8,b 3 x 8)=2 and we shall derive a contradiction. Then by (4.162) we obtain that
$$\bar{b}_3x_8=b_{10}+\alpha_{14}\quad\hbox{where } \alpha_{14}\in B.$$
Now by (4.165), (4.194) and (4.287) we obtain that
$$S+b_3e_9=t+b_{10}+\alpha_{14}.$$
Since 12≤|s| we obtain by (4.287) that |S|≤12 which implies that α 14Irr(b 3 e 9) and by (4.171) we obtain that (b 10,b 3 e 9)=1, which is a contradiction.
Now we have that (b 3 x 8,b 3 x 8)=3. Then by (4.162) we obtain that
$$\bar{b}_3x_8=b_{10}+\alpha+\beta\quad\hbox{where }\alpha,\beta\in B.$$
(4.289)
Now by (4.5) we obtain that b 8x 8, and by (4.288) we obtain that
$$b_3x_8=c_{15}+v_4+v_5\quad\hbox{where } v_4,v_5\in B.$$
(4.290)
Now by (4.165), (4.194) and (4.287) we obtain that
$$S+b_3e_9=b_{10}+\alpha+\beta+t.$$
Now we can assume that
$$b_3e_9=b_{10}+t+\beta$$
(4.291)
and
$$b_3c_{15}=b_{10}+y_{11}+s+\alpha$$
(4.292)
which implies that 5≤|α|, and by (4.289) we obtain that |α|≤10. Therefore
$$5\leq |\alpha|\leq 10.$$
Now (4.289) implies that
$$4\leq |\beta|\leq 9.$$
By (4.165), (4.291) and (4.292) we obtain that
$$14\leq |s|\leq 19$$
which implies that αs, αy 11 and sy 11. Since (b 3 x 8,b 3 x 8)=3 we obtain by (4.289) that αb 10. Now by (4.292) we obtain that (b 3 c 15,b 3 c 15)=4. By (4.279) and (4.288) we obtain that
$$\bar{b}_3c_{15}=b_{15}+x_7+x_8+f_{15}\quad\hbox{where } f_{15}\in B.$$
(4.293)
By (4.198), (4.290) and (4.292) we obtain that
$$\bar{b}_{15}+c_{15}+v_4+v_5+b_3f_{15}=G_{12}+\bar{b}_3s+\bar{b}_3\alpha.$$
(4.294)
By (4.196) we obtain that v 4Irr(G 12) and since 14≤|s| we have that $$v_{4}\notin \mathit{Irr}(\bar{b}_{3}s)$$. Therefore $$(v_{4},\bar{b}_{3}\alpha)=1$$ which implies that |α|≤9 and by (4.292) we obtain that $$(c_{15},\bar{b}_{3}\alpha)=1$$ which implies that 8≤|α|. Thus
$$8\leq |\alpha|\leq 9$$
(4.295)
which implies by (4.289) that
$$5\leq |\beta|\leq 6,$$
(4.296)
and by (4.292) we obtain that 15≤|s|≤16. By (4.162), (4.289), (4.290), (4.293) and $$b_{3}(\bar{b}_{3}x_{8})=\bar{b}_{3}(b_{3}x_{8})$$ we obtain that
$$f_{15}+\bar{b}_3v_4+\bar{b}_3v_5=b_3\alpha+b_3\beta.$$
By (4.289) we obtain that (x 8,b 3 α)=(x 8,b 3 β)=1 which implies by (4.296) that (f 15,b 3 β)=0 and (f 15,b 3 α)=1. So 9≤|α| and by (4.295), we obtain that α=α 9 which implies by (4.292) that s=s 15. By (4.285) we obtain that
$$\bar{b}_3s_{15}=c_{15}+\bar{b}_{15}+T_{15},$$
and by (4.294) we obtain that $$v_{4}+v_{5}+b_{3}f_{15}=G_{12}+T_{15}+\bar{b}_{3}\alpha_{9}$$. By (4.292) we obtain that $$(c_{15},\bar{b}_{3}\alpha_{9})=1$$ which implies that either b 3 f 15=G 12+T 15+c 15+h 3 where h 3B, which is impossible, or v 5Irr(G 12).
Now we have that v 5Irr(G 12). By (4.196) we obtain that
$$\bar{b}_3y_{11}=c_{15}+d_6+v_5+v_7\quad\hbox{where }v_7\in B$$
which implies that
$$b_3v_5=y_{11}+m_4\quad\hbox{where }m_4\in B,$$
and by (4.290) we obtain that
$$\bar{b}_3v_5=x_8+m_7\quad\hbox{where } m_7\in B.$$
Now by (4.290) and $$b_{3}(\bar{b}_{3}v_{5})=\bar{b}_{3}(b_{3}v_{5})$$ we obtain that $$v_{4}+b_{3}m_{7}=d_{6}+v_{7}+\bar{b}_{3}m_{4}$$. Thus
$$\bar{b}_3m_4=v_4+v_5+m_3\quad\hbox{where } m_3\in B$$
which implies that (m 3,b 3 m 7)=1, and we have a contradiction.

### 4.5.6 Case c=c16

In this section we assume that c=c 16. Then (4.195) and (4.197) imply that
$$\bar{b}_3c_{16}=x_7+b_{15}+H_{26}$$
(4.297)
and
$$b_3c_{16}=b_{10}+y_{11}+F_{27}.$$
(4.298)
By (4.169) and (4.174) we obtain that
$$\bar{b}_3d_5=x_7+g_{8}\quad\hbox{where } g_8\in B.$$
(4.299)
Assume henceforth that y 11 is a real element and we shall derive a contradiction. Then by (4.170) and (4.196) we obtain that
$$b_3y_{11}=x_7+\bar{d}_5+\bar{c}_{16}+h_5\quad\hbox{where } h_5\in B.$$
(4.300)
Now by (4.174) and (4.297) we obtain that
$$H_{26}+g_8=x_8+\bar{c}_{16}+\bar{d}_5+h_5.$$
Now we have that $$H_{26}=\bar{c}_{16}+\bar{d}_{5}+h_{5}$$ which implies by (4.297) that $$h_{5}\in \mathit{Irr}(\bar{b}_{3}c_{16})$$, which is a contradiction.
Now we have that y 11 is a non-real element. By (4.198) we obtain that
$$\bar{b}_{15}+b_3H_{26}=G_{12}+\bar{b}_3F_{27}$$
which implies that $$\bar{b}_{15}\in \mathit{Irr}(\bar{b}_{3}F_{27})$$. By (4.5) and (4.162) we obtain that b 8Irr(b 3 c 16) and $$\bar{b}_{10}\notin \mathit{Irr}(b_{3}c_{16})$$ and by (4.298) we can assume that
$$s\in \mathit{Irr}(F_{27}).$$
(4.301)
By (4.165) we obtain that $$(s,b_{3}\bar{b}_{15})=1$$ which implies that
$$\bar{b}_3s=\bar{b}_{15}+c_{16}+T\quad\hbox{where } T\in\mathbb{N}B$$
(4.302)
which implies that
$$12\leq |s|,$$
(4.303)
and by (4.298) we obtain that
$$b_3c_{16}=b_{10}+y_{11}+s+S\quad\hbox{where }S\in\mathbb{N}B.$$
(4.304)
Now by (4.2), (4.165) and $$b_{3}(\bar{b}_{3}s)=(b_{3}\bar{b}_{3})s$$ we obtain that
$$b_8s=b_8+b_{10}+\bar{b}_{10}+y_{11}+s+t+S+b_3T.$$
(4.305)
By (4.171) we obtain that
$$b_3e_8=b_{10}+\varSigma _{14}\quad\hbox{where}\varSigma _{14}\in\mathbb{N}B$$
(4.306)
and by (4.162) we obtain that
$$\bar{b}_3x_8=b_{10}+\varSigma ^*_{14}.$$
(4.307)
Now by (4.165) and (4.304) we obtain that
$$S+\varSigma _{14}=t+\varSigma ^*_{14}.$$
Now we have the following three cases:
1. Case 1:

S=t and $$\varSigma _{14}=\varSigma ^{*}_{14}$$.

2. Case 2:

$$S=\varSigma ^{*}_{14}$$ and Σ 14=t 14.

3. Case 3:

$$\varSigma ^{*}_{14}=\alpha+\beta$$ where α,βB, S=α and Σ 14=β+t.

Case 1 By (4.304) we obtain that
$$b_3c_{16}=b_{10}+y_{11}+s+t,$$
(4.308)
and by (4.165) we obtain that
$$\bar{b}_3t=\bar{b}_{15}+c_{16}+\varGamma \quad\hbox{where } \varGamma \in\mathbb{N}B$$
(4.309)
which implies that 12≤|t|. By (4.305) we obtain that
$$\bar{y}_{11}\in \mathit{Irr}(b_3T),$$
(4.310)
and by (4.302) we obtain that sIrr(b 3 T) which implies that 8≤|T| and 13≤|s|. In the same way we can show that
$$\bar{y}_{11}\in \mathit{Irr}(b_3\varGamma )$$
(4.311)
and 13≤|t|. Now by (4.165) we can assume that s=s 13 and t=t 14. Now (4.302) and (4.309) imply that
$$\bar{b}_3s_{13}=\bar{b}_{15}+c_{16}+v_8\quad\hbox{where }v_8\in B$$
and
$$\bar{b}_3t_{14}=\bar{b}_{15}+c_{16}+v_{11}\quad\hbox{where }v_{11}\in B.$$
By (4.170), (4.310) and (4.311) we obtain that
$$b_3y_{11}=x_7+\bar{v}_8+\bar{v}_{11}+f_7\quad\hbox{where }f_7\in B.$$
By (4.174), (4.297) and (4.299) we obtain that $$H_{26}+g_{8}=x_{8}+\bar{v}_{8}+\bar{v}_{11}+f_{7}$$, and together with (4.297) we obtain that $$(\bar{b}_{3}c_{16},\bar{b}_{3}c_{16})=5$$, which is a contradiction to (4.308).
Case 2 By (4.304) we obtain that
$$b_3c_{16}=b_{10}+y_{11}+s_{13}+\varSigma ^*_{14},$$
by (4.306) we obtain that
$$b_3e_8=b_{10}+t_{14},$$
(4.312)
by (4.172) we obtain that
$$\bar{b}_3e_8=b_{15}+h_9\quad\hbox{where}h_9\in B$$
(4.313)
and by (4.302) we obtain that
$$\bar{b}_3s_{13}=\bar{b}_{15}+c_{16}+v_8\quad\hbox{where } v_8\in B.$$
(4.314)
Now (4.305) implies that $$\bar{y}_{11}\in \mathit{Irr}(b_{3}v_{8})$$. Therefore
$$b_3v_8=\bar{y}_{11}+s_{13}.$$
(4.315)
Now (4.174) implies that either $$(\bar{v}_{8},\bar{b}_{3}c_{16}=1)$$ or $$(\bar{v}_{8},\bar{b}_{3}d_{5})=1$$.
Assume henceforth that $$(\bar{v}_{8},\bar{b}_{3}d_{5})=1$$ and we shall derive a contradiction. Then by (4.169) we obtain that
$$\bar{b}_3d_5=x_7+\bar{v}_8$$
(4.316)
and by (4.315) we obtain that
$$\bar{b}_3v_8=\bar{d}_5+f_{19}\quad\hbox{where } f_{19}\in B.$$
Now by (4.314), (4.316) and $$b_{3}(\bar{b}_{3}v_{8})=\bar{b}_{3}(b_{3}v_{8})$$, we obtain that $$\bar{x}_{7}+b_{3}f_{19}=\bar{b}_{15}+c_{16}+\bar{b}_{3}\bar{y}_{11}$$. Hence $$(\bar{f}_{19},b_{3}b_{15})=1$$, which is a contradiction to (4.172).
Now we have that $$(\bar{v}_{8},\bar{b}_{3}c_{16})=1$$ which implies by (4.315) that
$$\bar{b}_3v_8=\bar{c}_{16}+\gamma_8\quad\hbox{where }\gamma_8\in B$$
(4.317)
and by (4.297) we obtain that
$$\bar{b}_3c_{16}=x_7+b_{15}+\bar{v}_8+\varSigma _{18}\quad\hbox{where }\varSigma _{18}\in\mathbb{N}B.$$
Now by (4.3), (4.5), (4.162), (4.165), (4.172), (4.313), (4.314) and $$b_{3}(\bar{b}_{3}b_{15})=\bar{b}_{3}(b_{3}b_{15})$$ we obtain that $$x_{8}+b_{3}t_{14}=\bar{e}_{8}+\varSigma _{18}+b_{15}+h_{9}$$, and by (4.312) we obtain that $$(\bar{e}_{8},b_{3}t_{14})=1$$. Therefore (x 8,Σ 18)=1 which implies that
$$\bar{b}_3c_{16}=x_7+b_{15}+\bar{v}_8+x_8+h_{10}\quad\hbox{where }h_{10}\in B,$$
and by (4.174) and (4.299) we obtain that
$$b_3y_{11}=x_7+g_8+\bar{v}_8+h_{10}.$$
(4.318)
Now by (4.314), (4.315), (4.317) and $$b_{3}(\bar{b}_{3}v_{8})=\bar{b}_{3}(b_{3}v_{8})$$ we obtain that $$c_{16}+v_{8}+\bar{g}_{8}=\bar{x}_{8}+b_{3}\gamma_{8}$$. By (4.317) we obtain that (v 8,b 3 γ 8)=1 which implies that g 8=x 8. By (4.318) we obtain that $$(y_{11},\bar{b}_{3}x_{8})=(x_{8},b_{3}y_{11})=1$$, and by (4.5) we obtain that b 8x 8. Now by ( 4.162 ) we obtain that $$(b_{10},\bar{b}_{3}x_{8})=1$$, which is a contradiction.
Case 3 By (4.306) we obtain that
$$b_3e_8=b_{10}+\beta+t$$
(4.319)
which implies that $$(e_{8},\bar{b}_{3}t)=1$$, and by (4.165) we obtain that $$(\bar{b}_{15},\bar{b}_{3}t)=1$$. Therefore 9≤|t| and |β|≤5. By (4.5) and (4.319) we obtain that b 8e 8. Therefore |β|≠3 which implies that 4≤|β|≤5.
Assume henceforth that β=β 4 and we shall derive a contradiction. Then (4.319) implies that
$$\bar{b}_3\beta_4=e_8+h_4\quad\hbox{where } h_4\in B.$$
By (4.307) we obtain that
$$\bar{b}_3x_8=b_{10}+\alpha_{10}+\beta_4$$
which implies that
$$b_3\beta_4=x_8+g_4\quad\hbox{where } g_4\in B.$$
Now (4.319) and $$b_{3}(\bar{b}_{3}\beta_{4})=\bar{b}_{3}(b_{3}\beta_{4})$$ imply that $$t_{10}+b_{3}h_{4}=\alpha_{10}+\bar{b}_{3}g_{4}$$ which implies that t 10=α 10. By (4.165) we obtain that (b 15,b 3 t 10)=1 and by (4.304) we obtain that $$(\bar{c}_{16},b_{3}t_{10})=1$$, which is a contradiction.
Now we have that β=β 5, and by (4.165) and (4.319) we obtain that t=t 9, s=s 18 and α=α 9. By (4.165), (4.172) and (4.319) we obtain that
$$\bar{b}_3t_9=\bar{b}_{15}+e_8+\alpha_4\quad\hbox{where }\alpha_4\in B$$
and
$$\bar{b}_3e_8=b_{15}+f_4+f_5\quad\hbox{where } f_4,f_5\in B.$$
Now by (4.5), (4.3) (4.162), (4.165), (4.297), (4.302), (4.172) and $$b_{3}(\bar{b}_{3}b_{15})=\bar{b}_{3}(b_{3}b_{15})$$, we obtain that $$x_{8}+\bar{\alpha}_{4}+\bar{T}_{23}=f_{5}+f_{4}+H_{26}$$ which implies that $$\bar{f}_{5}\in \mathit{Irr}(T_{23})$$, and we have a contradiction to (4.302).

## 4.6 Case (b3x7,b3x7)=3

Since (b 3 x 7,b 3 x 7)=3 we obtain that
$$b_3x_7=c+d+e\quad\hbox{where } c,d,e\in B,$$
(4.320)
and by (4.162) we obtain that
$$\bar{b}_3x_7=b_{10}+w+z\quad\hbox{where } w,z\in B.$$
(4.321)
Now by (4.162) and $$b_{3}(\bar{b}_{3}x_{7})=\bar{b}_{3}(b_{3}x_{7})$$ we obtain that
$$\bar{b}_3c+\bar{b}_3d+\bar{b}_3e=b_{15}+x_7+x_8+b_3z+b_3w.$$
(4.322)
Now we can assume that (c,b 3 b 15)=1. By (4.3) we obtain that (b 6,b 3 b 15)=1, by (4.10) we obtain that $$(\bar{b}_{15},b_{3}b_{15})=1$$ and (b 3 b 15,b 3 b 15)=4. Now since $$c\neq b_{6},\bar{b}_{15}$$ we obtain that
$$b_3b_{15}=b_6+\bar{b}_{15}+c+f\quad\hbox{where }f\in B,$$
(4.323)
and by (4.10) we obtain that
$$\bar{b}_3b_{10}=b_6+c+f.$$
(4.324)
By (4.320) we obtain that |c|≤13 and $$(\bar{b}_{3}c,x_{7})=1$$ and by (4.323) we obtain that $$(\bar{b}_{3}c,b_{15})=1$$ which implies that
$$9\leq |c|\leq 13.$$
(4.325)
Now (4.324) implies that
$$11\leq |f|\leq 15.$$
(4.326)
By (4.321) we obtain that either $$\bar{b}_{3}x_{7}=z_{4}+w_{7}$$ or $$\bar{b}_{3}x_{7}=z_{5}+w_{6}$$.
Assume henceforth that
$$\bar{b}_3x_7=b_{10}+z_4+w_7$$
and we shall derive a contradiction. Then
$$b_3z_4=x_7+g_5\quad\hbox{where } g_5\in B$$
and
$$\bar{b}_3z_4=\alpha+\beta\quad\hbox{where}\alpha,\beta\in B.$$
(4.327)
Now $$b_{3}(\bar{b}_{3}z_{4})=\bar{b}_{3}(b_{3}z_{4})$$ implies that
$$b_3\alpha+b_3\beta=b_{10}+z_4+w_7+\bar{b}_3g_5.$$
(4.328)
Now we can assume that
$$(b_{10},b_3\alpha)=1.$$
(4.329)
By (4.327) we obtain that |α|≤9, by (4.4) we obtain that αb 6 and by (4.326) we obtain that fα. Now we have by (4.324) that c=α, by (4.325) we obtain that c=c 9 and by (4.327) we obtain that β=β 3. By (4.328) we obtain that $$b_{3}c_{9}+b_{3}\beta_{3}=b_{10}+z_{4}+w_{7}+\bar{b}_{3}g_{5}$$ which implies that (w 7,b 3 c 9)=1. By (4.327) we obtain that (z 4,b 3 c 9)=1 and together with (4.329) we obtain that 4≤(b 3 c 9,b 3 c 9). By (4.320) and (4.323) we obtain that (b 3 c 9,b 3 c 9)≤3, and we have a contradiction.
Now we have that
$$\bar{b}_3x_7=b_{10}+z_5+w_6$$
(4.330)
and
$$b_3z_5=x_7+\varSigma _8\quad\hbox{where}\varSigma _8\in\mathbb{N}B.$$
(4.331)
Now $$b_{3}(\bar{b}_{3}z_{5})=\bar{b}_{3}(b_{3}z_{5})$$ implies that $$b_{3}(\bar{b}_{3}z_{5})=b_{10}+z_{5}+w_{6}+\bar{b}_{3}\varSigma _{8}$$, by (4.4) we obtain that $$(b_{6},\bar{b}_{3}z_{5})=0$$ and by (4.324) we obtain that
$$\hbox{ either }\quad (c,\bar{b}_3z_5)=1\quad\hbox{ or }\quad (f,\bar{b}_3z_5)=1.$$
(4.332)
By (4.331) we obtain that 2≤(b 3 z 5,b 3 z 5)≤3.
Assume henceforth that (b 3 z 5,b 3 z 5)=3 and we shall derive a contradiction. Then by (4.325), (4.326) and (4.332) we obtain that
$$\bar{b}_3z_5=c_9+m_3+n_3\quad\hbox{where } m_3,n_3\in B.$$
Now (4.330), (4.331) and $$b_{3}(\bar{b}_{3}z_{5})=\bar{b}_{3}(b_{3}z_{5})$$ imply that $$b_{3}c_{9}+b_{3}m_{3}+b_{3}n_{3}=b_{10}+z_{5}+w_{6}+\bar{b}_{3}\varSigma _{8}$$. Hence b 3 c 9=b 10+z 5+w 6+v 6 where v 6B. By (4.320) and (4.323) we obtain that (b 3 c 9,b 3 c 9)≤,3 and we have a contradiction. Now (4.331) implies that
$$b_3z_5=x_7+g_8\quad\hbox{where } g_8\in B.$$
(4.333)
Assume henceforth that c=c 13 and we shall derive a contradiction. Then by (4.324) we obtain that f=f 11 and (4.332) implies that
$$\bar{b}_3z_5=f_{11}+\gamma_4\quad\hbox{where }\gamma_4\in B.$$
By (4.320) we obtain that
$$b_3x_7=c_{13}+d_4+e_4,$$
$$\bar{b}_3d_4=x_7+q_5\quad\hbox{where } q_5\in B$$
and
$$\bar{b}_3e_4=x_7+u_5\quad\hbox{where } u_5\in B.$$
Now by (4.322) and (4.333) we obtain that $$\bar{b}_{3}c_{13}+q_{5}+u_{5}=b_{15}+x_{8}+g_{8}+b_{3}w_{6}$$ which implies that (q 5,b 3 w 6)=(u 5,b 3 w 6)=1, and by (4.330) we obtain that (x 7,b 3 w 6)=1, which is a contradiction. Now we have by (4.325) that 9≤|c|≤12 and by (4.324) we obtain that 12≤|f|≤15.
Assume henceforth that $$(f,\bar{b}_{3}z_{5})=1$$ and we shall derive a contradiction. Then f=f 12 and
$$\bar{b}_3z_5=f_{12}+\gamma_3\quad\hbox{where}\gamma_3\in B.$$
(4.334)
By (4.324) we obtain that c=c 12. By (4.1), (4.320), (4.330) and $$b_{3}(b_{3}x_{7})=b^{2}_{3}x_{7}$$ we obtain that b 3 c 12+b 3 d+b 3 e=b 10+z 5+w 6+b 6 x 7. By (4.320) and since c=c 12 we obtain that |d|,|e|≠3,12 which implies by (4.334) that (z 5,b 3 d)=(z 5,b 3 e)=0. Hence (z 5,b 3 c 12)=1 which implies that c 12=f 12 and since (b 3 b 10,b 3 b 10)=3 we have a contradiction to (4.324).
Now by (4.332) and (4.333) we obtain that
$$\bar{b}_3z_5=c+\delta\quad\hbox{where}\delta\in B.$$
(4.335)
By (4.324) we obtain that
$$b_3c=b_{10}+z_5+D\quad\hbox{where } D\in\mathbb{N}B,$$
(4.336)
and by (4.320) and (4.323) we obtain that
$$\bar{b}_3c=x_7+b_{15}+C\quad\hbox{where }C\in\mathbb{N}B.$$
(4.337)
Now by (4.320), (4.323), (4.324) and $$b_{3}(\bar{b}_{3}c)=\bar{b}_{3}(b_{3}c)$$, we obtain that $$d+e+\bar{b}_{15}+b_{3}C=\delta+\bar{b}_{3}D$$. By (4.335) we obtain that |δ|≠15 which implies that $$\delta\neq \bar{b}_{15}$$. Therefore $$\bar{b}_{15}\in \mathit{Irr}(\bar{b}_{3}D)$$. Since 9≤|c|≤12 we obtain by (4.5), (4.162) and (4.336) that b 8Irr(D) and $$\bar{b}_{10}\notin \mathit{Irr}(D)$$. Therefore by (4.165) we can assume that sIrr(D). Now (4.336) implies that
$$b_3c=b_{10}+z_5+s+E\quad\hbox{where } E\in\mathbb{N}B.$$
(4.338)
Now (4.165) implies that
$$\bar{b}_3s=\bar{b}_{15}+c+F\quad\hbox{where } F\in\mathbb{N}B.$$
By (4.2), (4.165) and $$(b_{3}\bar{b}_{3})s=b_{3}(\bar{b}_{3}s)$$ we obtain that
$$b_8s=b_{10}+z_5+E+b_8+\bar{b}_{10}+s+t+b_3F.$$
Since b 8 and s are real elements and z 5 is a non-real element, we obtain that either $$\bar{z}_{5}\in \mathit{Irr}(E)$$ or $$\bar{z}_{5}\in \mathit{Irr}(b_{3}F)$$. If $$\bar{z}_{5}\in \mathit{Irr}(E)$$ then by (4.338) we obtain that $$(\bar{c},b_{3}z_{5})=1$$, and since 9≤|c|≤12 we have a contradiction to (4.333). Therefore $$\bar{z}_{5}\in \mathit{Irr}(b_{3}F)$$. Now by (4.333) we obtain that either $$\bar{x}_{7}\in \mathit{Irr}(F)$$ or $$\bar{g}_{8}\in \mathit{Irr}(F)$$. If $$\bar{x}_{7}\in \mathit{Irr}(F)$$ then $$(s,\bar{b}_{3}x_{7})=(\bar{x}_{7},\bar{b}_{3}s)=1$$. Since s is a real element and b 10, z 5 are non-real elements we obtain that sb 10, z 5, so (4.330) implies that s=w 6, which is a contradiction to (4.165) and (4.330). Now we have that
$$\bar{b}_3s=\bar{b}_{15}+c+\bar{g}_8+G\quad\hbox{where } G\in\mathbb{N}B.$$
(4.339)
By (4.333) we obtain that
$$\bar{b}_3g_8=z_5+s+H\quad\hbox{where }H\in\mathbb{N}B$$
(4.340)
Assume henceforth that c=c 9 and we shall derive a contradiction. Then by (4.337) we obtain that
$$\bar{b}_3c_9=x_7+b_{15}+u_5\quad\hbox{where } u_5\in B,$$
and by (4.338) we obtain that
$$b_3c_9=b_{10}+z_5+s_{12}.$$
By (4.339) we obtain that
$$\bar{b}_3s_{12}=\bar{b}_{15}+c_9+\bar{g}_8+u_4\quad\hbox{where } u_4\in B$$
which implies that b 3 u 4=s 12. Now by (4.320), (4.323), (4.324), (4.335) and $$b_{3}(\bar{b}_{3}c_{9})=\bar{b}_{3}(b_{3}c_{9})$$ we obtain that $$d+e+b_{3}u_{5}=\delta_{6}+c_{9}+\bar{g}_{8}+u_{4}$$. Hence (b 3 u 4,b 3 u 4)≠1, and we have a contradiction.
Assume henceforth that c=c 10 and we shall derive a contradiction. Then by (4.337) we obtain that
$$\bar{b}_3c_{10}=x_7+b_{15}+u_8\quad\hbox{where } u_8\in B$$
which implies by (4.338) that
$$b_3c_{10}=b_{10}+z_5+s_{15}.$$
By (4.339) we obtain that
$$\bar{b}_3s_{15}=\bar{b}_{15}+c_{10}+\bar{g}_8+G_{12}.$$
Now by (4.320), (4.323), (4.324), (4.335) and $$b_{3}(\bar{b}_{3}c_{10})=\bar{b}_{3}(b_{3}c_{10})$$, we obtain that $$d+e+b_{3}u_{8}=\delta_{5}+c_{10}+\bar{g}_{8}+G_{12}$$. By (4.320) we obtain that |d|+|e|=11, and now we can assume that d=δ 5, e=e 6 and e 6Irr(G 12) which implies that
$$\bar{b}_3s_{15}=\bar{b}_{15}+c_{10}+\bar{g}_8+e_6+\alpha_6\quad\hbox{where }\alpha_6\in B$$
and
$$b_3e_6=s_{15}+\alpha_3\quad\hbox{where } \alpha_3\in B.$$
By (4.320) we obtain that
$$\bar{b}_3e_6=x_7+\alpha_{11}.$$
Now by (4.320) and $$b_{3}(\bar{b}_{3}e_{6})=\bar{b}_{3}(b_{3}e_{6})$$ we obtain that $$d_{5}+b_{3}\alpha_{11}=\bar{b}_{15}+\bar{g}_{8}+\alpha_{6}+\bar{b}_{3}\alpha_{3}$$ which implies that $$(\bar{\alpha}_{11},b_{3}b_{15})=1$$, which is a contradiction to (4.323).
Assume henceforth that c=c 11 and we shall derive a contradiction. Then by (4.335) we obtain that δ=δ 4 and (z 5,b 3 δ 4)=1 which implies that (w 6,b 3 δ 4)=0. By (4.330), (4.333), (4.335), (4.340), (4.338) and $$b_{3}(\bar{b}_{3}z_{5})=\bar{b}_{3}(b_{3}z_{5})$$, we obtain that w 6+z 5+H=E+b 3 δ 4. Now we have that w 6Irr(E) which implies that (w 6,b 3 c 11)=1, and by (4.338) we obtain that
$$b_3c_{11}=b_{10}+z_5+s+w_6+Q\quad\hbox{where } Q\in\mathbb{N}B.$$
Hence 4≤(b 3 c 11,b 3 c 11). Now (4.337) implies that
$$\bar{b}_3c_{11}=x_7+b_{15}+u_5+u_6\quad\hbox{where} u_5,u_6\in B.$$
By (4.320), (4.323), (4.324), (4.335) and $$b_{3}(\bar{b}_{3}c_{11})=\bar{b}_{3}(b_{3}c_{11})$$, we obtain that $$d+e+\bar{b}_{15}+b_{3}u_{5}+b_{3}u_{6}=\delta_{4}+\bar{b}_{3}s+\bar{b}_{3}w_{6}$$. By (4.339) we obtain that $$\bar{g}_{8}\in \mathit{Irr}(\bar{b}_{3}s)$$, and since (c 11,b 3 u 5)=(c 11,b 3 u 6)=1 we obtain that $$(\bar{g}_{8},b_{3}u_{5})=(\bar{g}_{8},\allowbreak b_{3}u_{6})=\nobreak 0$$. Therefore we have that either $$d=\bar{g}_{8}$$ or $$e=\bar{g}_{8}$$, but by (4.169) we obtain that |d|+|e|=10 and we have a contradiction.
Now we have that c=c 12 which implies by (4.335) that δ=δ 3 and (z 5,b 3 δ 3)=1 which implies that (w 6,b 3 δ 3)=0. By (4.330), (4.333), (4.335), (4.340), (4.338) and $$b_{3}(\bar{b}_{3}z_{5})=\bar{b}_{3}(b_{3}z_{5})$$, we obtain that w 6+z 5+H=E+b 3 δ 3. Now we have that w 6Irr(E) which implies that (w 6,b 3 c 12)=1, and by (4.338) we obtain that
$$b_3c_{12}=b_{10}+z_5+s+w_6+Q\quad\hbox{where } Q\in\mathbb{N}B$$
which implies that |s|≥15. If |s|=15 then By (4.320), (4.323), (4.324), (4.335) and $$b_{3}(\bar{b}_{3}c_{12})=\bar{b}_{3}(b_{3}c_{12})$$ we obtain that $$d+e+\bar{b}_{15}+b_{3}C_{14}=\delta_{3}+\bar{b}_{3}s_{15}+\bar{b}_{3}w_{6}$$. By (4.320) we obtain that d,eδ 3 and since c=c 12 we obtain that $$(d,\bar{b}_{3}s_{15})=(e,\bar{b}_{3}s_{15})=0$$. Now we have that $$(d,\bar{b}_{3}w_{6})=(e,\bar{b}_{3}w_{6})=1$$, which is a contradiction to $$(c_{12},\bar{b}_{3}w_{6})=1$$. If |s|≠15 then |s|≤11, which is a contradiction to (4.339).

## 4.7 Case b3b10=b15+x5+y5+z5

Since b 3 b 10=b 15+x 5+y 5+z 5 we obtain that
$$\bar{b}_3x_5=b_{10}+u_5\quad\hbox{where } u_5\in B$$
(4.341)
and
$$\bar{b}_3y_5=b_{10}+v_5\quad\hbox{where } v_5\in B.$$
(4.342)
Now we have that 1≤(b 3 x 5,b 3 y 5) which implies that
$$b_3x_5=c+d\quad\hbox{where } c,d\in B$$
(4.343)
and
$$b_3y_5=c+e\quad\hbox{where } e\in B.$$
(4.344)
Assume henceforth that d=e and we shall derive a contradiction. Then (b 3 x 5,b 3 y 5)=2 and by (4.341) and (4.342), we obtain that
$$\bar{b}_3x_5=\bar{b}_3y_5=b_{10}+u_5$$
(4.345)
which implies that (x 5,b 3 u 5)=(y 5,b 3 u 5)=1. Now we have that
$$b_3u_5=x_5+y_5+g_5\quad\hbox{where } g_5\in B,$$
(4.346)
and by $$b_{3}(\bar{b}_{3}x_{5})=\bar{b}_{3}(b_{3}x_{5})$$ we obtain that
$$\bar{b}_3c+\bar{b}_3d=b_{15}+2x_5+2y_5+z_5+g_5$$
which implies that
$$|c|,|d|\geq 5$$
(4.347)
and we can assume that $$(b_{15},\bar{b}_{3}c)=1$$. By (4.343) and (4.344) we obtain that $$(x_{5},\bar{b}_{3}c)=(y_{5},\bar{b}_{3}c)=1$$ which implies that |c|≥10 and by (4.343) and (4.347) we obtain that c=c 10 and d=d 5. Since (c 10,b 3 b 15)=1 we obtain by (4.10) that $$(c_{10},\bar{b}_{3}b_{10})=1$$ which implies by (4.4) that
$$\bar{b}_3b_{10}=b_6+c_{10}+\alpha+\beta\quad\hbox{where }\alpha,\beta\in B.$$
(4.348)
Now by (4.1), (4.341), (4.343) and $$\bar{b}_{3}(\bar{b}_{3}x_{5})=\bar{b}^{2}_{3}x_{5}$$ we obtain that $$\bar{b}_{3}b_{10}+\bar{b}_{3}u_{5}=d_{5}+c_{10}+\bar{b}_{6}x_{5}$$. If $$(d_{5},\bar{b}_{3}b_{10})=1$$ then by (4.10) we obtain that (d 5,b 3 b 15)=1, and we have a contradiction.
Now we have that
$$(d_5,\bar{b}_3b_{10})=0$$
(4.349)
which implies that $$(d_{5},\bar{b}_{3}u_{5})=1$$ and by (4.346) we obtain that (b 3 u 5,b 3 u 5)=3 which implies that
$$\bar{b}_3u_5=d_5+\gamma+\delta\quad\hbox{where }\gamma,\delta\in B.$$
(4.350)
By (4.345), (4.346) and $$b_{3}(\bar{b}_{3}u_{5})=\bar{b}_{3}(b_{3}u_{5})$$ we obtain that $$b_{3}d_{5}+b_{3}\gamma+b_{3}\delta=2b_{10}+2u_{5}+\bar{b}_{3}g_{5}$$. By (4.349) we obtain that (b 10,b 3 γ)=(b 10,b 3 δ)=1. By (4.350) and (4.4) we obtain that γ,δb 6 and |γ|+|δ|=10 which implies that γ,δc 10. Now (4.348) implies that γ+δ=α+β and |α|+|β|=14, which is a contradiction.
Now we have that $$(\bar{b}_{3}x_{5},\bar{b}_{3}y_{5})=1$$. In the same way, we can show that $$(\bar{b}_{3}x_{5},\bar{b}_{3}z_{5})=(\bar{b}_{3}y_{5},\bar{b}_{3}z_{5})=1$$ and we obtain that
$$\bar{b}_3z_5=b_{10}+w_5,$$
(4.351)
and by (4.343) and (4.344) we obtain that either b 3 z 5=d+e or b 3 z 5=c+f where fB. If b 3 z 5=d+e then 2|c|=15, so
$$b_3z_5=c+f\quad\hbox{where } f\in B.$$
(4.352)
By $$b_{3}(\bar{b}_{3}x_{5})=\bar{b}_{3}(b_{3}x_{5})$$, $$b_{3}(\bar{b}_{3}y_{5})=\bar{b}_{3}(b_{3}y_{5})$$ and $$b_{3}(\bar{b}_{3}z_{5})=\bar{b}_{3}(b_{3}z_{5})$$, we obtain that
$$\begin{array}{c}\bar{b}_3c+\bar{b}_3d=b_{15}+x_5+y_5+z_5+b_3u_5,\\[6pt]\bar{b}_3c+\bar{b}_3e=b_{15}+x_5+y_5+z_5+b_3v_5\end{array}$$
(4.353)
and
$$\bar{b}_3c+\bar{b}_3f=b_{15}+x_5+y_5+z_5+b_3w_5.$$
By $$\bar{b}_{3}(\bar{b}_{3}x_{5})=\bar{b}^{2}_{3}x_{5}$$ we obtain that
$$\bar{b}_3b_{10}+\bar{b}_3u_5=c+d+\bar{b}_6x_5.$$
(4.354)
Assume henceforth that (c,b 3 b 15)=0 and we shall derive a contradiction. Then (d,b 3 b 15)=(e,b 3 b 15)=(f,b 3 b 15)=1. In addition, we have that |d|=|e|=|f| which implies by (4.4) and (4.10) that
$$\bar{b}_3b_{10}=b_6+d_8+e_8+f_8.$$
Now by (4.352) we obtain that c=c 7 and by (4.354) we obtain that $$(c_{7},\allowbreak \bar{b}_{3}u_{5})=\nobreak1$$. By (4.4), (4.341), (4.342), (4.351) and $$b_{3}(\bar{b}_{3}b_{10}=\bar{b}_{3}(b_{3}b_{10})$$, we obtain that $$b_{8}+b_{3}d_{8}+b_{3}e_{8}+b_{3}f_{8}=\bar{b}_{3}b_{15}+2b_{10}+u_{5}+v_{5}+w_{5}$$ which implies that $$(\bar{b}_{3}u_{5},\allowbreak\bar{b}_{3}u_{5})=\nobreak 2$$. By (4.353) and since (c,b 3 b 15)=0 we obtain that $$(b_{15},\bar{b}_{3}d_{8})=1$$ and together with (4.343) we obtain that
$$\bar{b}_3d_8=b_{15}+x_5+g_4\quad\hbox{where } g_4\in B,$$
and by (4.353) we obtain that (g 4,b 3 u 5)=1. Now by (4.341) and $$(\bar{b}_{3}u_{5},\bar{b}_{3}u_{5})=2$$ we obtain that b 3 u 5=g 4+x 5, which is a contradiction.
Now we have that $$(b_{15},\bar{b}_{3}c)=1$$ and by (4.10) we obtain that $$(c,\bar{b}_{3}b_{10})=1$$ which implies that
$$\bar{b}_3b_{10}=b_6+c+\alpha+\beta\quad\hbox{where } \alpha,\beta\in B$$
(4.355)
and
$$b_3b_{15}=\bar{b}_{15}+b_6+c+\alpha+\beta.$$
(4.356)
Now we have that
$$|\alpha|,|\beta|\geq 6.$$
(4.357)
By (4.343), (4.344) and (4.352) we obtain that $$(x_{5},\bar{b}_{3}c)=(y_{5},\bar{b}_{3}c)=(z_{5},\bar{b}_{3}c)=1$$, and together with $$(b_{15},\bar{b}_{3}c)=1$$ and b 3 x 5=c+d we obtain that either c=c 10 or c=c 12. By (4.341) we obtain that (x 5,b 3 b 10)=(x 5,b 3 u 5)=1 which implies that
$$(\bar{b}_3u_5,\bar{b}_3b_{10})\geq 1.$$
(4.358)
Assume henceforth that c=c 12 and we shall derive a contradiction. Then by (4.356) we obtain that α=α 6, β=β 6 and
$$\bar{b}_3\alpha_6=b_{15}+h_3\quad\hbox{where } h_3\in B.$$
(4.359)
By (4.354) and (4.356) we obtain that $$b_{6}+c_{12}+\alpha_{6}+\beta_{6}+\bar{b}_{3}u_{5}=c_{12}+d_{3}+\bar{b}_{6}x_{5}$$ which implies that
$$(d_3,\bar{b}_3u_5)=1.$$
(4.360)
By (4.4) we obtain that $$(b_{6},\bar{b}_{3}u_{5})=0$$. By (4.355) we obtain that (b 10,b 3 α 6)=1 and by (4.359) we obtain that (b 3 α 6,b 3 α 6)=2 which implies that $$0=(u_{5},b_{3}\alpha_{6})=(\alpha_{6},\bar{b}_{3}u_{5})$$. In the same way, we can show that $$(\beta_{6},\bar{b}_{3}u_{5})=0$$ which implies by (4.358) and (4.360) that $$\bar{b}_{3}u_{5}=d_{3}+c_{12}$$. Now we have that (b 3 u 5,b 3 u 5)=2 and by (4.341) we obtain that
$$b_3u_5=x_5+g_{10}\quad\hbox{where } g_{10}\in B.$$
By (4.353) we obtain that $$\bar{b}_{3}c_{12}+\bar{b}_{3}d_{3}=b_{15}+2x_{5}+y_{5}+z_{5}+g_{10}$$, and we have a contradiction.
Now we have that c=c 10 which implies that
$$\bar{b}_3c_{10}=b_{15}+x_5+y_5+z_5$$
(4.361)
and by (4.354) and (4.355) we obtain that $$b_{6}+\alpha+\beta+\bar{b}_{3}u_{5}=d_{5}+\bar{b}_{6}x_{5}$$. By (4.357) we obtain that α,βd 5 which implies that
$$(d_5,\bar{b}_3u_5)=1.$$
(4.362)
By (4.4) we obtain that $$(b_{6},\bar{b}_{3}u_{5})=0$$. If $$(\alpha,\bar{b}_{3}u_{5})=1$$ then $$(\bar{b}_{6}x_{5},\bar{b}_{6}x_{5})\geq 7$$. By (4.2) and (4.343) we obtain that $$(b_{8},x_{5}\bar{x}_{5})=1$$ in addition $$(b_{8},b_{6}\bar{b}_{6})=1$$. Now we have that
$$x_5\bar{x}_5=1+b_8+\varSigma _{16}\quad\hbox{where }\varSigma _{16}\in\mathbb{N}B$$
and
$$b_6\bar{b}_6=1+b_8+\varSigma _{27}\quad\hbox{where }\varSigma _{27}\in\mathbb{N}B.$$
By (4.12) we obtain that $$b_{3} \bar{b}_{15}=b_{8}+\bar{b}_{10}+\varSigma _{27}$$ and since (b 3 b 15,b 3 b 15)=5 we have a contradiction. Now (4.358) implies that (u 5,b 3 c 10)=1. In the same way, we can show that (v 5,b 3 c 10)=(w 5,b 3 c 10)=1 and by (4.355) we obtain that (b 10,b 3 c 10)=1 which implies that (b 3 c 10,b 3 c 10)≠4, and we have a contradiction to (4.361).

© Springer-Verlag London Limited 2011

## Authors and Affiliations

• 1
• 2
Email author
• Xu Bangteng
• 3
• Guiyun Chen
• 4
• Effi Cohen
• 1
• Arisha Haj Ihia Hussam
• 5
• Mikhail Muzychuk
• 6
1. 1.Department of MathematicsBar Ilan UniversityRamat GanIsrael
2. 2.Netanya Academic CollegeNetanyaIsrael
3. 3.Department of Mathematics and StatisticsEastern Kentucky UniversityRichmondUSA
4. 4.Department of MathematicsSouthwest UniversityChongqingPeople’s Republic of China
5. 5.Department of MathematicsAlqasemi Academic College of EducationBaqa El-GharbiehIsrael
6. 6.Netanya Academic CollegeNetanyaIsrael