Twin Composites, Strange Continued Fractions, and a Transformation that Euler Missed (Twice)

Abstract

We introduce a polynomial E(d, t, x) in three variables that comes from the intersections of a family of ellipses described by Euler. For fixed odd integers $t\ge 3$, the sequence of E(d, t, x) with d running through the integers produces, conjecturally, sequences of “twin composites” analogous to the twin primes of the integers. This polynomial and its lower degree relative R(d, t, x) have strikingly simple discriminants and resolvents. Moreover, the roots of R for certain values of d have continued fractions with at least two large partial quotients, the second of which mysteriously involves the 12th cyclotomic polynomial. Various related polynomials whose roots also have conjecturally strange continued fractions are also examined.

Appendix: More About E

Here, we provide more details about the nature of $E=E(d, t, x)$ and its relation to $R=R (d, t,x)$.

Theorem 7.3

Let $1<t<d$. Then, all roots of the quartic

$$\begin{aligned} E (d, t, \sqrt{x})= 0 \end{aligned}$$

are real and positive.

Proof

Inspection of the unwieldy solution by radicals formula shows that all roots are real. For $d=t$, the discriminant is 0, and we have

$$\begin{aligned} E (t, t, x)= t^4 (1-4 t^2 x + 4 t^2 x^2)^2. \end{aligned}$$

The roots of the above are all positive, since $t>1$. It has two distinct roots, each of multiplicity two. As d increases from t, these multiple roots immediately separate into distinct positive roots lying on the real axis, since the roots must remain real. Since the discriminant does not change sign for d in the open interval $(t, \infty )$, the roots remain henceforth distinct and real. If, for some $d>t$, one of them becomes negative, then, by continuity, one of them became 0 for a smaller value of d. However

$$\begin{aligned} E (d, t, 0)= t^4, \end{aligned}$$

so this cannot happen. The result follows. $\square$

Remark 7.4

One can show by implicit differentiation that as d increases from t, each of the two double roots separates horizontally into two simple roots at a rate that is initially infinite. However, the ratio of the rate of separation of the larger root to that of the smaller root is the larger root of

$$\begin{aligned} 1-(4t^2 - 1) x + x^2 = 0. \end{aligned}$$

We now establish the relationship between E and R.

Theorem 7.5

If we arrange the roots $s_i$ of $E(d, t, \sqrt{x})$, so that $0 \le s_1 \le s_2 \le s_3 \le s_4,$ then the positive square root of $s_3/s_2$ is a root of R(d, t, x).

Proof

Proceed as in Sect. 5. In $E(d, t, \sqrt{x})$, replace x by $\frac{tx}{(2 d^2)}$ and call the new roots the $s_i'$. This change of variable results in a self-reciprocal polynomial, so $s_1' s_4' = s_2' s_3' = 1$. Thus, for the original roots, we have

$$\begin{aligned} s_1 s_4 = s_2 s_3 = t^2 / (4 d^4). \end{aligned}$$

We now apply the cubic resolvent of Sect. 3 much as we did with a cubic resolvent in Sect. 5. We obtain

$$\begin{aligned} s_1 s_2 + s_3 s_4&= \frac{t^2 (2 d^2 - 1)}{2 d^4},\\ s_1 s_3 + s_2 s_4&= \frac{2 d^2 - t^2}{2 d^4}, \end{aligned}$$

and

$$\begin{aligned} s_1 s_4 + s_2 s_3 = \frac{t^2}{2 d^4}. \end{aligned}$$

Division of the second of the above equations by $s_2 s_3 = s_1 s_4$ yields

$$\begin{aligned} (s_1 / s_2) + (s_2 / s_1) = 2(2 d^2 - t^2)/t^2, \end{aligned}$$

and similarly, the third yields

$$\begin{aligned} (s_1 / s_3) + (s_3 / s_1) = 2(2 d^2 - 1). \end{aligned}$$

Now, $x + (1/x)=r$ is equivalent to $x^2-rx+1=0$ and a simple calculation with resultants shows that if also $y^2 - sy +1 = 0,$ then xy is a root of $T = T (r, s, z)$, where

$$\begin{aligned} T = 1 - rsz + (r^2 + s^2 -2) z2 - rsz^3 + z^4. \end{aligned}$$

Thus, to find a multiple of the minimal polynomial of

$$\begin{aligned} (s_3 / s_2) = (s_3 / s_1) (s_1 / s_2), \end{aligned}$$

we simply use the values of r and s provided by the above equations. Computer algebra then yields the factorization

$$\begin{aligned} T (r, s, z^2) = R (d, t, z) R (d, t, -z). \end{aligned}$$

Thus, the positive square root of the ratio is a root of R(d, t, z) and the result follows. $\square$