A number theory problem by Furkan Nane

This is a partial solution only, but it should be straightforward to fix up. The main idea is that if you squint at the equation, you see that $n$ should be about $m^2$ .

If we write $n = m^2+a$ , we get a quartic in $m$ that looks like $(3a-5)m^4+(3a^2-10a)m^2+m+(a^3-5a^2) = 0$ . If $a > 5$ then all the coefficients are positive, so there are no nonnegative roots. There should be a way to get a lower bound on negative values of $a$ as well, maybe for sufficiently large $m$ . Perhaps someone can fill this in.

In practice I just looked at all the quartics for $-5 \le a \le 5$ , and got one solution each when $a = -1, a = 0, a = 2, a = 4, a = 5$ . The solutions are $(1,0), (0,0), (3,11), (1,5),$ and $(0,5)$ . So there are $\fbox{5}$ in all.