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Theorem 1.18

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Theorem
NumberTheory

Theorem

If pp is a prime divisor of bn1b^{n}-1 then either 1. p(bd1)p|(b^{d}-1) for some dnd|n or 2. p1(modn)p\equiv 1\pmod{n} If p>2p>2 and nn odd, then for 2, we have p1(mod2n)p\equiv 1\pmod{2n}.