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October 12[edit]

Congruence question[edit]

How could it be proven that if p is a prime number and n an integer such that 2^n-1 ≡ 1 (mod n) with p²| n implies 2^p-1 ≡ 1 (mod p²)? I understand that from p² | n it follows that 2^n-1 ≡ 1 (mod p²), but how does 2^p-1 ≡ 1 (mod p²) follow from that? --Toshio Yamaguchi (tlk−ctb) 09:04, 12 October 2012 (UTC)[reply]

Ah, n-1 is divisible by the multiplicative order of 2 modulo p². -- Toshio Yamaguchi (tlk−ctb) 10:06, 12 October 2012 (UTC)[reply]

Okay, then I have a follow up question: How do I know that p-1 is a multiple of ord_p² 2? -- Toshio Yamaguchi (tlk−ctb) 10:18, 12 October 2012 (UTC)[reply]

By the Euler–Fermat theorem, ord_p²(2) divides φ(p²) = (p − 1)p, hence also gcd(n − 1,(p −1)p). Since n − 1 is coprime to p, this gcd divides p − 1.—Emil J. 14:22, 12 October 2012 (UTC)[reply]