Let $n \geq 1$ be some integer. Can we always find a prime power $p^k$ such that $p^k - 1$ has exactly $n$ distinct prime divisors?
For example:
- $n = 1$ example: $2^2 - 1 = 3$
- $n = 2$ example: $5^2 - 1 = 2^2 \cdot 3$
- $n = 3$ example: $7^3 - 1 = 2 \cdot 3^2 \cdot 19$
- $n = 4$ example: $5^6 - 1 = 2^3 \cdot 3^2 \cdot 7 \cdot 31$
We can always find a prime power $p^k$ with $\geq n$ prime divisors. Let $p_1, p_2, \ldots, p_n$ be distinct primes, $p_i \neq p$ for each $i$. Then $p^k \equiv 1 \mod{p_1p_2 \ldots p_n}$ for some positive $k$, for example we can choose $$k = (p_1 - 1)(p_2 - 1)\ldots(p_n - 1)$$