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Can $9xy$ divide $1+x^2+x^3+y^2$ for integers $x,y$? Equivalently, do there exist integers $x,y,z$ such that $$ 1 + x^2 + x^3 + y^2 + 9 x y z = 0 \quad ? $$ This equation arises in the search for the shortest cubic equation with open Hilbert 10-th problem, see e.g. On the shortest open cubic equation

I searched for $|y|$ up to 100 millions and $|x|$ up to 300 millions, with no solutions found.

It is easy to see that $y \equiv 3 \bmod 6$, while $x \equiv 4 \bmod 9$ and is not divisible by $4$.

Michael Hardy
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Bogdan Grechuk
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  • I now searched for |y| up to 100 millions and |x| up to 300 millions, with no solutions found. The question is updated to include this information. – Bogdan Grechuk Mar 06 '24 at 09:17
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    Consider a simpler problem: $xy \mid x^3 + x^d + 1 + y^2$, $d \in {1, 2}$. If $(x_0, y_0)$ is solution, then $(x_0, \frac{x_0^3 + x_0^d + 1}{y_0}, d)$ and $(\frac{y_0^2 + 1}{x_0}, y_0, 3 - d)$ are solutions as well. We can use this to find new solutions, but the size of solutions grows very fast. – Denis Shatrov Mar 06 '24 at 18:01

1 Answers1

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The equation is solvable in integers. Take, for example, $$ x = -19578556686240310295378317903565, \\ y = -101658411567714319887, \\ z = 418962851513108789978912616277180591709694. $$

Verification can be done by substitution.

I found this solution by using the transformations described by Denis Shatrov in the comment. We consider equation $$ 1 + x^2 + x^3 + y^2 + x y z = 0 \quad\quad (1) $$ and look for a solution such that $z$ is divisible by $9$. If we start with any solution $(x_0,y_0,z_0)$, then, as observed by Denis, $$ (x_1,y_1,z_1)=\left(\frac{y_0^2+1}{x_0}, y_0, -\frac{1+x_0+x_0^3+y_0^2}{x_0y_0}\right) $$ solves equation $$ 1 + x + x^3 + y^2 + x y z = 0 \quad\quad (2). $$ Then $$ (x_2,y_2,z_2)=\left(x_1, \frac{x_1^3+x_1+1}{y_1}, -\frac{1+x_1+x_1^3+y_1^2}{x_1y_1}\right) $$ is also a solution to (2), while $$ (x_3,y_3,z_3)=\left(\frac{y_2^2+1}{x_2}, y_2, -\frac{1+x_2^2+x_2^3+y_2^2}{x_2y_2}\right) $$ is again a solution to (1).

By doing modulo 9 analysis, I observed that if $(x_0,y_0,z_0)$ is $(4,0,3)$ modulo $9$, then $(x_3,y_3,z_3)$ is $(4,6,0)$ modulo $9$. An easy computer search returned solution $(x_0,y_0,z_0) = (-3965, 1446687, 354)$ to (1) which is $(4,0,3)$ modulo $9$. Then the corresponding $(x_3,y_3,z_3)$ is a solution to (1) with $z$ divisible by (9), hence $(x_3,y_3,z_3/9)$ is an integer solution to the original equation $1 + x^2 + x^3 + y^2 + 9x y z = 0$. As mentioned above, its correctness can be easily verified by direct substitution.

Peter Mueller
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Bogdan Grechuk
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  • Here is another solution obtained the same way, with positive $x$: $$(x,y) = (134398663297274547209137686278055005569690302475018, 99289900600732241944365446997153).$$ – Max Alekseyev Mar 20 '24 at 20:12