Find all strictly positive numbers $a, p, n$ such that $p^{a}-1=$ $2^{n}(p-1)$, where $p$ is a prime number.
It is clear that $p>2$. Suppose by contradiction that $a=$ $uv$ is composite. Then according to Zsigmondy's theorem, $p^{u}-1$ has a prime factor $q$ that does not divide $p-1$. But $p^{u}-1$ divides $p^{a}-1=2^{n}(p-1)$. Therefore, $q=2$. However, $p-1$ is even, which is a contradiction. Therefore, $a$ is prime.

If $a=2$, we find that $p=2^{n}-1$.

If $a>2$, similarly, Zsigmondy's theorem implies that $2^{n}(p-1)=p^{a}-1$ has a prime factor $r$ that does not divide $p-1$. This implies that $r=2$, which is absurd because $p-1$ is even.

The solutions are therefore $a=2$ and $n$ such that $2^{n}-1$ is prime.