Mathematical Reflections 2013, Issue 3 - Problem J269

Problem:
Solve in positive integers the equation $$(x^2-y^2)^2-6\min(x,y)=2013.$$

Proposed by Titu Andreescu.

Solution:
Clearly, $x \neq y$. Suppose without loss of generality that $x<y$. Then, $$2013+6x=(x-y)^2(x+y)^2>(x+y)^2>4x^2,$$ which gives $4x^2-6x-2013<0$. Hence, $0<x<23$. Moreover, $$(x^2-y^2)^2=3(671+2x),$$ therefore $671+2x$ must be divisible by $3$ since the left member is a perfect square. This implies that $x=3k+2$ for some $k \in \mathbb{N}$, so $$(x^2-y^2)^2=9(225+2k),$$ and $225+2k$ must be a perfect square. If $k=0$ it's obvious. The least positive integer such that $225+2k$ is a square is $k=32$, but for this value we get $x=98>23$. Therefore $k=0$, $x=2$ and $(x^2-y^2)^2=2025=45^2$ which gives $y^2-x^2=45$, i.e. $y=7$. By symmetry, $x=7,y=2$ is another solution of the equation. So, all the positive integer solutions of the given equation are $(2,7),(7,2)$.