Saudi Arabian Mathematical Competitions, 2015

Notice that if $x < 0$ or $y < 0$ then the term $2^{x} 5^{y}$ is not an integer while $x^{2} y^{5} - 2015 - 4 x y$ is an integer which is impossible. If $x = 0$ or $y = 0$ then the equation becomes $-2^{x} 5^{y} = 2015$ which has no solution. We deduce that both $x, y$ are positive integers. Moreover, because $2^{x} 5^{y} + 2015 + 4 x y$ is an odd integer, both $x, y$ are odd integers.

Working modulo $5$, we get $x^{2} y \equiv 4 x y$ and therefore either $x \equiv 0, 4$ or $y \equiv 0 \pmod{5}$.

Working modulo $8$, and because $x, y$ are odd, we get $y - 2^{x} \cdot 5 \equiv -1 + 4$ which is equivalent to $y \equiv 5(2^{x} - 1) \pmod{8}$. Therefore, either $x = 1$ and $y \geq 5$ or $x \geq 3$ and $y \equiv 3 \pmod{8}$.

If $x = 1$ and $y \geq 5$ then the equation becomes $y^{5} - 2 \cdot 5^{y} = 2015 + 4 y$ and has no solution since the left hand side is negative (We prove it by induction on $y \geq 5$).

If $x = 3$ then $y \equiv 3 \pmod{8}$. But $y \equiv 0 \pmod{5}$. Therefore $y \geq 35$ and $9 \cdot y^{5} - 8 \cdot 5^{y}$ is negative so the equation has no solution.

If $x \geq 5$ and $y \neq 3$ then $y \geq 5$. In this case $x^{2} < 2^{x}$ and $y^{5} \leq 5^{y}$ (We prove it by induction) and therefore $x^{2} y^{5} - 2^{x} 5^{y}$ is negative and the equation has no solution.

If $x \geq 7$ and $y = 3$ then because $x \equiv 0, 4 \pmod{5}$, $x \geq 9$. Therefore $x^{2} \cdot 3^{5} - 2^{x} 5^{3} < 3^{5}(x^{2} - 2^{x-2})$ is negative and therefore the equation has no solution.

If $x = 5$ and $y = 3$, the equation is satisfied.

Hence the unique solution is $(5, 3)$.