Problem

Prove $x^2 + y^2 = x^2y^2$ has no non-trivial integer solutions.

Solution (by Willow, from US)

Assume for the sake of contradiction that there exists $x,y \in \mathbb{Z}, x,y \neq 0$ such that $x^2 + y^2 = x^2y^2$. Rearranging the equation we have

$$ \begin{align*} \\y^2 &= x^2y^2 - x^2 \\y^2 &= x^2(y^2-1)\\ \frac{y^2}{x^2} &= y^2 -1 \\ \frac{y}{x} &= \sqrt{y^2 -1} \end{align*} $$

Because $y$ and $x$ are integers, $\frac yx$ is rational. This implies that $\sqrt{y^2 -1}$ is rational. This is a contradiction as one less than a non trivial square number is non square, and the square root of a non square integer is irrational.

Thank you!

Special thanks to Philip Yao, Artem (Ukraine), Kei Beauduin (Palaiseau, France), Simone (Italia), Enrico M. (Italy), Marcos Lopez (Wichita Falls, Tx), khushii, Anthony (Seattle, WA), Ajit C. Balram (Chennai), Jason Kenny (San Diego), Andrew (Folsom, California), Shyaam ganesh (India), Provatas Fotis (Greece), Willow (US), Isaac (Hong Kong), Jakub Waniek (UK), Jaron Bailey (Australia), Ankit (Bay Area), and Max (New York) for submitting a solution to this challenge problem!