Mongolian Mathematical Olympiad

Let $s=m+n$ and $p=mn$. Then the equation becomes: $$s^2+p+1=Ns$$ So, $$p=Ns-s^2-1$$ Recall that $m$ and $n$ are roots of the quadratic equation $x^2-sx+p=0$.

The discriminant is: $$D=s^2-4p=s^2-4(Ns-s^2-1)=s^2-4Ns+4s^2+4$$ $$=5s^2-4Ns+4$$ For $m$ and $n$ to be integers, $D$ must be a perfect square.

Let $D=t^2$ for some integer $t$. So, $$5s^2-4Ns+4=t^2$$ This is a quadratic in $s$ for each integer $t$.

For each integer $t$, the equation $$5s^2-4Ns+4-t^2=0$$ has discriminant in $s$: $$\Delta =(4N{)}^2-4\cdot 5\cdot (4-t^2)=16N^2-20(4-t^2)=16N^2-80+20t^2$$ For sufficiently large $|t|$, this is positive, so there are integer solutions for $s$ for infinitely many $t$.

Alternatively, fix $s$ to be any integer, then $p=Ns-s^2-1$ is also integer, and $m$ and $n$ are roots of $x^2-sx+p=0$.

The roots are $$m,n=\frac{s\pm \sqrt{s^2-4p}}{2}$$ So $s^2-4p$ must be a perfect square. But $s^2-4p=s^2-4(Ns-s^2-1)=5s^2-4Ns+4$

So for each $s$ such that $5s^2-4Ns+4$ is a perfect square, we get integer solutions.

But for each $k$, let $s$ be such that $5s^2-4Ns+4=k^2$ for some integer $k$. This is a quadratic in $s$: $$5s^2-4Ns+4-k^2=0$$ For each integer $k$, this quadratic has integer solutions for $s$ for infinitely many $k$.

Therefore, there are infinitely many integer solutions $(m,n)$ to the given equation for any integer $N$.