36. When $p=2$, $a^{n}=13$, we know $a=13, n=1$. When $p>2$, since $p$ is a prime, $p$ is an odd number, at this time
$$2^{p}+3^{p}=(2+3)\left(2^{p-1}-2^{p-2} \times 3+\cdots-2 \times 3^{p-2}+3^{p-1}\right),$$

thus $5 \mid a^{n}$, which means $5 \mid a$. If $n>1$, then $5^{2} \mid a^{n}$, in this case, we should have
$$2^{p-1}-2^{p-2} \times 3+\cdots-2 \times 3^{p-2}+3^{p-1} \equiv 0(\bmod 5)$$

Using $3 \equiv-2(\bmod 5)$, $p$ is an odd number, and the above equation, we know
$$\begin{aligned} & 2^{p-1}-2^{p-2} \times 3+\cdots-2 \times 3^{p-2}+3^{p-1} \\ \equiv & \underbrace{2^{p-1}+2^{p-1}+\cdots+2^{p-1}}_{p \uparrow 2^{p-1}} \\ = & p \cdot 2^{p-1} \equiv 0(\bmod 5) \end{aligned}$$

Therefore, $5 \mid p$, and since $p$ is a prime, $p=5$, which leads to $a^{n}=2^{5}+3^{5}=275=5^{2} \times 11, n$ can only be 1, a contradiction.

Thus, $n=1$.