Slovenia — National Math Olympiad, 2012

Denote $\sqrt[3]{n} + \frac{p}{\sqrt[3]{n}} = k^2$ where $k$ is a natural number. We raise the equation to the 3rd power and get $n + 3p\sqrt[3]{n} + 3\frac{p^2}{\sqrt[3]{n}} + \frac{p^3}{n} = k^6$, which is $n + 3pk^2 + \frac{p^3}{n} = k^6$. From this we see that $n$ must divide $p^3$. Since $p$ is prime, we conclude $n = 1$, $n = p$, $n = p^2$ or $n = p^3$. If $n = p$ or $n = p^2$, we get the equation $p + 3pk^2 + p^2 = k^6$. Hence $k$ must be divisible by $p$, so the right side of the equation is divisible by $p^2$, but the left is not. We still have to check $n = 1$ and $n = p^3$. If we substitute them into the equation, we get $1 + p = k^2$, hence $p = (k-1)(k+1)$. From this we conclude that $k=2$ and $p=3$. We get two solutions: $n = 1$ and $p = 3$, $n = 27$ and $p = 3$.