Junior Balkan MO

Solution:
Suppose that the positive integer $n$ is such that
$$p^{n}-(p-1)^{n}=3^{a}$$
for some prime $p$ and positive integer $a$.

If $p=2$, then $2^{n}-1=3^{a}$ by $(1)$, whence $(-1)^{n}-1 \equiv 0\pmod{3}$, so $n$ should be even. Setting $n=2s$ we obtain $\left(2^{s}-1\right)\left(2^{s}+1\right)=3^{a}$. It follows that $2^{s}-1$ and $2^{s}+1$ are both powers of $3$, but since they are both odd, they are co-prime, and we have $2^{s}-1=1$, i.e. $s=1$ and $n=2$.

If $p=3$, then (1) gives $3 \mid 2^{n}$, which is impossible.

Let $p \geq 5$. Then it follows from (1) that we can not have $3 \mid p-1$. This means that $2^{n}-1 \equiv 0$ $(\bmod 3)$, so $n$ should be even, and let $n=2k$. Then
$$p^{2k}-(p-1)^{2k}=3^{a} \Longleftrightarrow \left(p^{k}-(p-1)^{k}\right)\left(p^{k}+(p-1)^{k}\right)=3^{a}$$
If $d=\left(p^{k}-(p-1)^{k}, p^{k}+(p-1)^{k}\right)$, then $d \mid 2p^{k}$. However, both numbers are powers of $3$, so $d=1$ and $p^{k}-(p-1)^{k}=1$, $p^{k}+(p-1)^{k}=3^{a}$.

If $k=1$, then $n=2$ and we can take $p=5$. For $k \geq 2$ we have $1=p^{k}-(p-1)^{k} \geq p^{2}-(p-1)^{2}$ (this inequality is equivalent to $p^{2}\left(p^{k-2}-1\right) \geq (p-1)^{2}\left((p-1)^{k-2}-1\right)$, which is obviously true). Then $1 \geq p^{2}-(p-1)^{2}=2p-1 \geq 9$, which is absurd.

It follows that the only solution is $n=2$.