, 2012

Denote $\sqrt[n]{n + \frac{2}{p}} = k$ where $k$ is a natural number, hence $n + \frac{2}{p} = k^n$. Thus $p$ is a divisor of $2$. Since $p$ is prime, we have $p = 2$ or $p = 1$ (but $1$ is not prime). So $p = 2$.

Now, $n + \frac{2}{2} = k^n \implies n + 1 = k^n$.

Try $n = 1$: $1 + 1 = 2 = k^1 \implies k = 2$.
Try $n = 2$: $2 + 1 = 3 = k^2 \implies k = \sqrt{3}$ (not integer).
Try $n = 3$: $3 + 1 = 4 = k^3 \implies k = \sqrt[3]{4}$ (not integer).

So only $n = 1$, $p = 2$, $k = 2$ is a solution.

Alternatively, if the original problem was $\sqrt[n]{n + \frac{2}{n}}$ (as in the solution context), then $n$ must be a divisor of $2$, so $n = 1$ or $n = 2$.

For $n = 1$: $1 + \frac{2}{1} = 3 = k^1 \implies k = 3$.
For $n = 2$: $2 + \frac{2}{2} = 3 = k^2 \implies k = \sqrt{3}$ (not integer).

Thus, the only solution is $n = 1$, $p = 2$, $k = 2$.

Therefore, the solutions are $n = 1$, $p = 2$.