This is strange, since if I'm not wrong it follows from the following IMO 2001 shortlisted problem: If p>3 is a prime then we can find n<p-1 such that $ p^2$ doesn't divide $ n^{p-1}-1 $ nor $ (n+1)^{p-1}-1$. Now, suppose for a certain x $ p^2$ doesn't divide $ x^{p-1}-1$. It's quite obvious that there must be a prime q|x such that $ p^2$ does not divide $ q^{p-1}-1$. Thus we can find two different prime factors of n and n+1 with the property stated. They will be certainly different. Am I missing something?

Valiowk, you guys should really be studying the past ISLs right?

Well it isn't necessary to use the other shortlisted problem. It's enough to consider $(p-1)^{p-1}, (p+1)^{p-1}, (2p-1)^{p-1}, (p+1)^{p-1}$ and by a simple argument it can be shown that there must exist $q_1, q_2$ which are factors of some of $p-1, p+1, 2p-1, 2p+1$. And yes, Valentin, we should have studied the past ISLs harder...

The main observation is that if $x^{p-1} \not \equiv 1$ (mod $p^2$), then some prime divisor $q$ of $x$ satisfies $q^{p-1} \not \equiv 1$ (mod $p^2$). Indeed, if this were not the case, then we could multiply some numbers which are all $1$ (mod $p^2$) to obtain $x^{p-1}$, absurd. Now, with the cases $p < 11$ being trivial, we will use this observation to solve the cases where $p\ge 11.$ Call a prime $q$ tasty if $q^{p-1} \not \equiv 1$ (mod $p^2$). Notice that $(p-1)^{(p-1)} \equiv p+1$ (mod $p^2$) and $(p+1)^{(p-1)} \equiv -p+1$ (mod $p^2$). Since these are both not $1$ (mod $p^2$), both $p-1$ and $p+1$ have a prime divisor which is tasty. If these are not both $2$, we're done since they then must be distinct. Otherwise, we have that $2$ is tasty. Now, we have that $(2p-1)^{(p-1)} \equiv -2p(p-1) + 1 \not \equiv 1$ (mod $p^2$). This means that $2p-1$ has a tasty prime divisor, which is clearly in $(1, p-1)$ and not equal to $2.$ Together with $2$, we are done. $\square$