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From: http://oeis.org/A036236
a(1) = 0, that is, no n exists with 2^n mod n = 1. Proof. Assume that there exists such n > 1. Consider its smallest prime divisor p. Then 2^n == 1 (mod p) implying that the multiplicative order ord_p(2) divides n. However, since ord_p(2) < p and p is the smallest divisor of n, we have ord_p(2) = 1, that is, p divides 2^1 - 1 = 1 which is impossible. [From Max Alekseyev] |
