eigenvalues of integral matrices

Solution 1:

There is no solution to your problem. The pattern of proof is familiar : we show that there cannot be a solution with too large parameters, and then a finite numbers of cases remain to be checked by exhaustive inspection.

If a solution matrix exists, it has eigenvalues $a,a,\frac{1}{a^2}$ for some real $a\not\in\lbrace -1,0,1\rbrace$. So its characteristic polynomial is

$$ \chi= (X-a)^2(X-\frac{1}{a^2})= X^3-\frac{2a^3+1}{a^2}X^2+\frac{a^3+2}{a}X-1 \tag{1} $$

Then $x=\frac{2a^3+1}{a^2}$ and $y=\frac{a^3+2}{a}$ must be integers. Eliminating $a$, we deduce purely algebraically (computing resultants) that $Q(x,y)=0$ where

$$ Q(x,y)=4y^3 - x^2y^2 - 18xy + 4x^3 + 27 \tag{2} $$

Lemma 1. Suppose that $x \geq 17$. Then the univariate polynomial $Q(x,.)$ has exactly three real roots $\rho_1 < \rho_2 < \rho_3 $ with $\rho_i\in (\alpha_i,\beta_i)$, where $$ \begin{array}{lcl} \alpha_1=-\sqrt{4x-0.99} &,& \ \beta_1=-\sqrt{4x}, \\ \alpha_2=\sqrt{4x-0.99}&,& \ \beta_2=\sqrt{4x}, \\ \alpha_3=\frac{x^2}{4}&,& \ \beta_3=\frac{x^2+0.99}{4}\\ \end{array} $$

Corollary 2 (of Lemma 1). There is no integer solution of $Q(x,y)=0$ with $x\geq 17$.

Lemma 3. Suppose that $x \leq -8$. Then the univariate polynomial $Q(x,.)$ has exactly one real root $\rho \in (\alpha,\beta)$ where $\alpha=\frac{x^2-0.999}{2}$ and $\beta=\frac{x^2}{2}$.

Corollary 4 (of Lemma 3). There is no integer solution of $Q(x,y)=0$ with $x\leq -8$.

Once we have corollaries 2 and 4, all that remains to be done is check the cases $x=(-7),(-6),(-5), \ldots ,8$ one by one. Inspection reveals that for these values, $Q(x,.)$ has rational roots only when $x=(-1),3$ or $5$. For $x=5$, the rational root is $\frac{17}{4}$, a noninteger. Too bad ...

Proof of Lemma 1. If $x$ is $\geq 17$, the number $|\alpha_1|$ is $\geq \sqrt{4\times 17+0.99}\geq 8.1$, so $a=|\alpha_1|-8.1$ is nonnegative. Also, the number $\alpha_2$ is $\geq \sqrt{4\times 17-0.99}\geq 8.15$, so $b=\alpha_2-8.15$ is nonnegative. Also, $\beta_3 \geq \frac{17^2+0.99}{4} \times 72$. Finally, $c=x-17$ is nonnegative.

$$ \begin{array}{lclc} Q(x,\alpha_1)&=&-\frac{99}{1600}a^4 - \frac{6019}{4000}a^3 - \frac{30213}{2500}a^2 - \frac{7117389}{200000}a - \frac{6956739}{4000000} & <0 \\ & & & \\ Q(x,\beta_1)&=&\frac{|\beta_1|^3}{2}+27 & >0 \\ & & & \\ Q(x,\alpha_2)&=& \frac{99}{1600}b^4 + \frac{12137}{8000}b^3 + \frac{4018197}{320000}b^2 + \frac{20410041}{640000}b + \frac{12365891}{10240000} & >0 \\ & & & \\ Q(x,\beta_2)&=&-\frac{\beta_2^3}{2}+27 & <0 \\ & & & \\ Q(x,\alpha_3)&=& -\frac{x^3}{2}+27 & <0 \\ & & & \\ Q(x,\beta_3)&=& \frac{99}{1600}c^4 + \frac{1483}{400}c^3 + \frac{6553101}{80000}c^2 + \frac{31287117}{40000}c + \frac{43169498099}{16000000} & >0 \\ & & & \\ \end{array} $$

The intermediate value property then yields three roots for our degree three polynomial, and concludes the proof of the lemma.

Proof of Corollary 2. If there is an integer solution $(x,y)$, then $y$ must be one of $\rho_1,\rho_2,\rho_3$. If, for example, $y=\rho_1$, then $y^2-4x$ in an integer in $(0,1)$, which is impossible. The other cases are similar.

Proof of Lemma 3.

If $x$ is $\leq -8$, the number $d=|x|-8$ is nonnegative. Then

$$ \begin{array}{lclc} Q(x,\alpha)&=&-\frac{999}{16000}d^4 - \frac{749}{500}d^3 - \frac{94809999}{8000000}d^2 - \frac{17185749}{500000}d - \frac{12580874999}{16000000000} & <0 \\ & & & \\ Q(x,\beta)&=&\frac{|x|^3}{2}+27 & >0 \\ & & & \\ \end{array} $$

So we have a root $\rho\in(\alpha,\beta)$. Computing $$\frac{Q(x,y)}{y-\rho}= 4y^2+(4\rho-x^2)y+(4\rho^2-18x-\rho x^2) $$ and the discriminant of this trinomial in $y$ is $x^4+8\rho x^2 + 288x - 48\rho^2 < 0$, so $\rho$ is the only root.

Proof of Corollary 4. This is similar to, and simpler than, the proof of corollary 2.

Solution 2:

As in Ewan's proof, write $$x = \frac{2 a^3+1}{a^2} \quad y = \frac{a^3+2}{a}.$$ Notice that $$a = \frac{xy-9}{2 (x^2-3y)}. \quad (\ast)$$ So, if $x$ and $y$ are integers then either $a$ is rational or else $xy=9$ and $x^2=3y$. The latter case implies $x (x^2/3) = 9$, so $x=y=3$ and $a=1$, which we already ruled out.

So we focus on the case where $a$ is rational. Write $a=p/q$ in lowest terms. So $$x = \frac{2p^3+q^3}{p^2 q} \quad y=\frac{p^3+2 q^3}{p q^2}.$$ From the formula for $x$, any prime that divides $p$ also divides $q$; from the formula $y$, any prime that divides $q$ also divides $p$. Since $p/q$ is in lowest terms, we get $a = \pm 1$.

So, where did $(\ast)$ come from? Since the map $a \mapsto \left( \frac{2a^3+1}{a^2}, \frac{a^3+2}{a} \right)$ is generically injective, the subfield of $k(a)$ generated by $(2a^3+1)/a^2$ and $(a^3+2)/a$ should be all of $k(a)$. So such a formula should exist, and I just needed to find it. To find it, I used that $a$ is double root of $t^3-x t^2+y t - 1$, so it is also a root of $3 t^2 - 2x t + y$. Therefore, $a$ is a root of $(t^3-x t^2+y t - 1) - (t/3 - x/9) (3 t^2 - 2x t + y) = (2y/3 - 2x^2/9) t +(xy/9 -1)$. I solved this equation for $t$.