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Find all pairs of positive integers $(x, y)$ such that $9x^2+y$ and $9y^2+x$ are both perfect squares.

My process:

Supposed, $9x^2+y=a^2$ and $9y^2+x=b^2$

Then, $8(x+y)(x-y)=(a+b)(a-b)$

I don't know how to continue further. Any clue would be helpful.

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  • $\begingroup$ Check your algebra. $9(x+y)(x-y)-(x-y)=(9(x+y)-1)(x-y)$ $\endgroup$ – Cheerful Parsnip Jun 11 at 16:14
  • $\begingroup$ $$(9x^2+y)-(9y^2+x)=9(x^2-y^2)-(x-y)=(x-y)(9x+9y-1).$$ $\endgroup$ – Servaes Jun 11 at 16:14
  • $\begingroup$ Note that you must have $9y^2+x \geq (3y+1)^2$, ie $x > 6y > y$, and similarly $y > x$. $\endgroup$ – Mindlack Jun 11 at 16:39
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There is no such pair.

$9x^2+y$ a perfect square; $y$ positive; implies $9x^2+y \ge (3x+1)^2$ which implies $y \geq 6x+1$ which implies (for $x$ nonegative) that $y>x$. Likewise we derive $x > y$. No pair $(x,y)$ of numbers satisfy both simultaneously.

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