Question 112
Let $p$ be an odd prime. Let $a_1, a_2, \dots, a_{p-1}$ be integers such that $i \cdot a_i \equiv 1 \pmod{p}$, for all $i = 1, 2, \dots, p-1$. Prove that
$$
2^p - 2 \equiv p(a_1 - a_2 + \dots + a_{p-2} - a_{p-1}) \pmod{p^2}.
$$
$$
2^p - 2 \equiv p(a_1 - a_2 + \dots + a_{p-2} - a_{p-1}) \pmod{p^2}.
$$