# Preliminary Classification of Sub-case (3)

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)
$$\bar{b}_6x_5=\bar{b}_3b_{10}.$$
(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
$$H_{17}+\bar{b}_3d_8=x_8+\bar{c}_{13}+\bar{d}_8+x_7+h_5,$$
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)
$$2\leq (b_3x_8,b_3x_8)\leq 3.$$
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).

## 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