Wikipedia:Reference desk/Archives/Mathematics/2012 October 12

= October 12 =

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

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

Okay, then I have a follow up question: How do I know that p-1 is a multiple of ordp2 2? -- Toshio Yamaguchi (tlk−ctb) 10:18, 12 October 2012 (UTC)
 * By the Euler–Fermat theorem, ordp2(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)