Find the smallest prime $p>100$ for which there exists an integer $a>1$ such that $p$ divides $\frac{a^{89}-1}{a-1}$.
The answer is $p=179$. To see this works, take $a=4$; by Fermat's little theorem, $4^{89}-1=2^{178}-1$ is divisible by 179 .

Now suppose $a^{89} \equiv 1(\bmod p)$. We consider two cases:

- If $a \equiv 1(\bmod p)$, then

$$
0 \equiv 1+a+\cdots+a^{88} \equiv 89 \quad(\bmod p)
$$

which forces $p=89$.

- Otherwise, since 89 is prime, it follows $a$ has order 89 modulo $p$. So $89 \mid p-1$. The smallest prime which obeys this is $p=179$.