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If ppp is a prime divisor of bn−1b^{n}-1bn−1 then either 1. p∣(bd−1)p|(b^{d}-1)p∣(bd−1) for some d∣nd|nd∣n or 2. p≡1(modn)p\equiv 1\pmod{n}p≡1(modn) If p>2p>2p>2 and nnn odd, then for 2, we have p≡1(mod2n)p\equiv 1\pmod{2n}p≡1(mod2n).