31st Turkish Mathematical Olympiad

Let $p\equiv 3(\mathrm{mod}4)$ be a prime number. We will show that for $k=p^2$ there are no positive integers $m$ and $n$ satisfying the equation. Suppose that there is a solution. If we rewrite the equation as $$n^2-p^{2}n+(m^2-p^2{m}^4)=0$$ we obtain a quadratic equation for $n$ and the discriminant must be a perfect square. Therefore, $$p^4+4p^2{m}^4-4m^2=s^2$$ for some integer $s$. Then $p\mid (2m{)}^2+s^2$ and we get $m=px$ and $s=pt$ for some integers $x$ and $t$. So we have $$p^2+4p^4{x}^4-4x^2=t^2$$ and again we obtain $p\mid (2x{)}^2+t^2$. Thus, we have $x=py$ and $t=pu$ for some integers $y$ and $u$. Dividing both sides by $p^2$ gives $$1+4p^6{y}^4-4y^2=u^2$$ Let $z=2p^3$. Since $y\ne 0$ and $z>2$, we conclude $$(z{y}^2-1{)}^2=z^2{y}^4-2z{y}^2+1<z^2{y}^4-4y^2+1=u^2<z^2{y}^4=(z{y}^2{)}^2.$$ Since $u^2$ lies between two consecutive perfect squares, we get a contradiction.