, 2012

Denote $\sqrt{n} + \frac{p}{\sqrt{n}} = k^2$ where $k$ is a natural number.

Squaring both sides of the equation gives us
$$n + 2p + \frac{p^2}{n} = k^4.$$
Hence $n$ must divide $p^2$. Since $p$ is prime, we conclude $n = 1$, $n = p$ or $n = p^2$.

If $n = p$, we get the equation $p + 2p + p = k^4$ or $4p = k^4$. Hence $k$ must be even and $p$ must be divisible by $4$, which is impossible.

If $n = 1$ or $n = p^2$, we get the equation $1 + 2p + p^2 = k^4$ or $1 + p = k^2$, hence $p = (k-1)(k+1)$.

We conclude that $k=2$, hence $p=3$. This gives us two solutions: $n=1$ and $p=3$, $n=9$ and $p=3$.