2001 IMO Shortlist S21

Let $p \geq 5$ be a prime number. Prove that there exists an integer $a$ with $1 \leq a \leq p-2$ such that neither $a^{p-1}-1$ nor $(a+1)^{p-1}-1$ is divisible by $p^2$ .

设 $p \geq 5$ 为素数。证明存在一个整数 $a$ 且 $1 \leq a \leq p-2$ 使得 $a^{p-1}-1$ 和 $(a+1)^{p-1}-1$ 都不能被 $p^2$ 整除。