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In general, are there strategies for finding all integers $x$ and $y$ such that $x \mid p(y)$ and $y \mid p(x)$ for some polynomial $p$ with integer coefficients? For example, could we find all integers $x$ and $y$ (WLOG let $x \leq y$) such that $x \mid 2y^2-2y+1$ and $y \mid 2x^2-2x+1$. I know there are nontrivial solutions for this example: $(1, 1)$, $(17, 109)$, and $(29, 125)$ all satisfy these conditions, although I can't prove that there are or aren't any more solutions than that. Trying to express these conditions in equation form (i.e. $2y^2-2y+1 = kx$, $2x^2-2x+1 = my$) and eliminating variables doesn't seem to get me very far.

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  • $\begingroup$ If $p$ is quadratic and $xy\mid p(x)p(y)$ $\iff$ $x\mid p(y)$ and $y\mid p(x)$, it is helpful to rewrite it as $xy\mid p(x)p(y)$. (One example is $p(x)=x^2-1$.) After cancelling the $x^2y^2,x^2y,xy^2$ terms this gives a biquadratic in $x,y$ (with a parameter) which can, in theory, be solved (think of Pell's equation etc) for every value of the parameter. I think things will be much more complicated when $\deg p>2$. $\endgroup$ – Bart Michels Jun 14 '15 at 9:23
  • $\begingroup$ I'm not quite sure I understand. If we let $p(x) = x^2-1$, then $p(x)\cdot p(y) = x^2y^2-x^2-y^2+1$. Then, with parameter $m$, we have the equation $x^2y^2-x^2-mxy-y^2+1 = 0$. I don't see what substitution to make in order to rewrite this in the form $u^2-nv^2-1 = 0$ (Pell's equation). $\endgroup$ – Michael L. Jun 14 '15 at 18:00
  • $\begingroup$ The $x^2y^2$ term can be absorbed in $m$. $\endgroup$ – Bart Michels Jun 14 '15 at 18:17
  • $\begingroup$ True, if we let $m = xy$, then we get Pell's equation, but that removes a lot of the generality from the problem. Actually, in the case $m = xy$, there are no solutions (rather, the only solution is $(0, 1)$ or $(1, 0)$, which is extraneous, because a number generally can't be divisible by zero). $\endgroup$ – Michael L. Jun 14 '15 at 19:05
  • $\begingroup$ As an example of this, letting $m = xy$ throws away the solution $x = 2$, $y = 3$ (which comes from $m = 4$). Clearly, $2\mid 3^2-1 = 8$ and $3\mid 2^2-1 = 3$. $\endgroup$ – Michael L. Jul 30 '17 at 5:17

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