Winter Mathematical Competition

Let $(x^2+y)(y^2+x)=p^5$, where $p$ is a prime. Then $x^2+y=p^s$, $y^2+x=p^t$, where $\{s,t\}=\{1,4\}$ or $\{2,3\}$. In the first case we can assume without loss of generality that $x<y$, $x^2+y=p$ and $y^2+x=p^4$. Then $p^2=(x^2+y{)}^2>x+y^2=p^4$, a contradiction. Let $x<y$, $x^2+y=p^2$ and $y^2+x=p^3$. Note that $p>x$. We have $p^2|(x^2+y)(x^2-y)+(y^2+x)=x^4+x=x(x+1)(x^2-x+1)$ and since $p>x$ we see that $p^2$ divides $(x+1)(x^2-x+1)$. We consider two cases.

Case 1. If $p|x+1$ then $p=x+1$ and we easily find the solution $x=2,y=5$.

Case 2. If $p\nmid x+1$ then $p^2|x^2-x+1$ and now $p^2|x^2+y=(x^2-x+1)+(x+y-1)$ implies that $p^2$ divides $x+y-1$. Hence $y\ge p^2-x+1>p^2-p$ and $p^3=y^2+x>p^2(p-1{)}^2$ which is impossible.

Finally, the solutions are (2, 5) and (5, 2).