To supplement existing data, solutions of a(p-1) equivalent to 1 (mod 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) equivalent to 1 (mod 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) equivalent to 1 (mod p(r)) are generally considered for any r greater than or equal to 2 and fixed prime p to see where the smallest prime solution a occurs.