Saudi Arabian IMO Booklet

First observe that $n=k^2-1$ for some positive integer $k$. As $n$ is not a perfect square, $\tau (n)=k$ must be even, ergo $n$ is odd. We now establish a bound on $\tau (n)$. Recall that, since $n$ is not a perfect square, there exists a bijective correspondence between factors of $n$ greater than $\lfloor \sqrt{n}\rfloor =k-1$ and factors of $n$ at most $\lfloor \sqrt{n}\rfloor$. Furthermore, since $n$ is odd, all divisors of $n$ must also be odd. Therefore $$k=\tau (n)\le 2\cdot |\{1,3,\dots ,k-1\}|=2\cdot \frac{k}{2}=k.$$ Hence equality must hold, and so every odd number between 1 and $k-1$ must divide $n=k^2-1$. In particular, $k-3$ must divide $k^2-1$; but $$\frac{k^2-1}{k-3}=k+3+\frac{8}{k-3}.$$ Therefore, $k-3$ is either $-1$ or $1$, so $k$ equals either $2$ or $4$ and $n\in \{3,15\}$, both of these work. $\square$