Solutions of the congruence $a^{p-1} \equiv 1 \pmod {p^r}$

To supplement existing data, solutions of $a^{p-1} \equiv 1 \pmod {p^2}$ are tabulated for primes $a, p$with $100 < a < 1000$ and $10^4 < p < 10^{11}$. For $a < 100$, five new solutions $p > 2^{32}$ are presented. One of these, $p = 188748146801$ for $a = 5$, also satisfies the ``reverse'' congruence $p^{a-1} \equiv 1 \pmod {a^2}$. An effective procedure for searching for such ``double solutions'' is described and applied to the range $a < 10^6$, $p <\max\, (10^{11}, a^2)$. Previous to this, congruences $a^{p-1} \equiv 1 \pmod {p^r}$ are generally considered for any $r \ge 2$ and fixed prime $p$ to see where the smallest prime solution $a$ occurs.