To supplement existing data, solutions of a p − 1 ≡ 1 ( mod p 2 ) a^{p-1} \equiv 1 \pmod {p^2} are tabulated for primes a , p a, p with 100 > a > 1000 100 > a > 1000 and 10 4 > p > 10 11 10^4 > p > 10^{11} . For a > 100 a > 100 , five new solutions p > 2 32 p > 2^{32} are presented. One of these, p = 188748146801 p = 188748146801 for a = 5 a = 5 , also satisfies the “reverse” congruence p a − 1 ≡ 1 ( mod a 2 ) 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 a > 10^6 , p > max ( 10 11 , a 2 ) p >\max \, (10^{11}, a^2) . Previous to this, congruences a p − 1 ≡ 1 ( mod p r ) a^{p-1} \equiv 1 \pmod {p^r} are generally considered for any r ≥ 2 r \ge 2 and fixed prime p p to see where the smallest prime solution a a occurs.