## Solution.

If $p=2$, then $2^{p}+3^{p}=13$ and we would have $n=1$, which contradicts the premise of the problem.

Assume that $p>2$ and let $2^{p}+3^{p}=a^{n}$ for some natural number $n>1$.

Since $p$ is an odd number, we have

$$a^{n}=2^{p}+3^{p}=(2+3)\left(2^{p-1}-2^{p-2} \cdot 3+\cdots-2 \cdot 3^{p-2}+3^{p-1}\right)$$

so $a$ is divisible by 5. Therefore, the number on the right side is divisible by $5^{n}$. Since $n>1$, the number is divisible (at least) by 25, so we conclude that the expression in the second parenthesis on the right side must be divisible by 5.

Given that $3 \equiv-2(\bmod 5)$, we have:

$2^{p-1}-2^{p-2} \cdot 3+\cdots-2 \cdot 3^{p-2}+3^{p-1} \equiv 2^{p-1}+2^{p-2} \cdot 2+\cdots+2 \cdot 2^{p-2}+2^{p-1} \equiv p \cdot 2^{p-1} \quad(\bmod 5)$.

It follows that $p$ is divisible by 5, and since $p$ is a prime number, the only possibility left is $p=5$. However,

$$2^{5}+3^{5}=275=5^{2} \cdot 11$$

which is not a number of the form $a^{n}$ for $n>1$.