Argentine National Olympiad 2015

We check directly that $p=2$ is a solution $(2^3-4\cdot 2+9=3^2)$, and $p=3$ is not. Henceforth let $p>3$. If $p^3-4p+9=n^2$ for some $n\in \mathbb{N}$ then $p^3-4p=n^2-9$, $$(p-2)p(p+2)=(n-3)(n+3).$$ One of the numbers $p-2$, $p$, $p+2$ is divisible by $3$, hence $n$ is also a multiple of $3$. If $n=3k$, $k\ge 1$, then $$(p-2)p(p+2)=9(k-1)(k+1).$$ Both sides are nonzero as $p\ne 2$. The prime $p>3$ divides $9(k-1)(k+1)$, hence it divides exactly one of $k-1$ and $k+1$. Let $p$ divide $k-1$. Write $k-1=lp$ to obtain $$(p-2)(p+2)=9l(lp+2).$$ Reducing mod $p$ gives $18l+4\equiv 0(\mathrm{mod}p)$. Since $p$ is odd, it divides $9l+2$. In particular $p\le 9l+2$, $p-2\le 9l$, so $(p-2)(p+2)=9l(lp+2)$ implies $p+2\ge lp+2$. This is possible only if $l=1$ and $p=9l+2=11$.

We have a solution $p=11$ $(11^3-4\cdot 11+9=36^2)$. Let $p$ divide $k+1$ and $k+1=lp$. Then $$(p-2)(p+2)=9l(lp-2).$$ Take this mod $p$ to obtain $18l-4\equiv 0(\mathrm{mod}p)$. It follows that $p$ divides $9l-2$. In particular $p\le 9l-2$, $p+2\le 9l$, so $(p-2)(p+2)=9l(lp-2)$ implies $p-2\ge lp-2$. Therefore $l=1$ and $p=9l-2=7$. We find one more solution: $p=7$ $(7^3-4\cdot 7+9=18^2)$.

In summary, the solutions are $p=2,7,11$.