Jörg Richstein

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To supplement existing data, solutions of a p−1 ≡ 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 ≡ 1 (mod a 2). An effective procedure for searching for such " double solutions(More)
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