thanks for helpin me. guys(Bow)

1. let p be a prime, show that every prime divisor of (2^p) -1 is greater than p.

2. let n= 3^(t-1). show that 2^n = -1 (mod 3^t)

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- December 2nd 2008, 03:25 AMfelixmcgradynumber theory
thanks for helpin me. guys(Bow)

1. let p be a prime, show that every prime divisor of (2^p) -1 is greater than p.

2. let n= 3^(t-1). show that 2^n = -1 (mod 3^t) - December 2nd 2008, 04:39 AMPaulRS
**1.**Suppose is a prime divisor of then and thus, by Lagrange's Theorem, the order of divides . (1)

Suppose is the order of , it cannot be 1 for otherwise , and since the order must divide , but since is prime this implies that that is the order of is . So, by (1), thus thus

**2.**By induction, it's true for k=1, so let's assume it's true for some we'll show that this implies the assertion for

By hypothesis so now: ( see the terms in the binomial expansion), thus:

In fact 2 is a primitive root module for all - December 3rd 2008, 05:31 PMThePerfectHacker