Please help. :(

Let p be an odd prime. Given:

(x+1)^p=x^p + 1^p (mod p)

holds for any integer x, prove n^p=n(mod p) for all integers n.

Any help would be greatly appreciated!!

Printable View

- Nov 17th 2009, 07:42 PMzachschOdd Primes with Modulo
Please help. :(

Let p be an odd prime. Given:

(x+1)^p__=__x^p + 1^p (mod p)

holds for any integer x, prove n^p__=__n(mod p) for all integers n.

Any help would be greatly appreciated!! - Nov 17th 2009, 08:30 PMtonio
- Nov 17th 2009, 08:36 PMzachsch
How would I go about "taking it from here"?

- Nov 17th 2009, 08:51 PMtonio
- Nov 17th 2009, 09:25 PMzachsch
Got it! Thanks a million!!!

- Nov 18th 2009, 08:59 AMgmatt
By the way you said in your hypothesis that p has to be odd, implying that this does not work for p=2. It does in fact work for p=2.

Also, the proof of the hypothesis is far more interesting than the proof of your statement, which is practically a corollary of the hypothesis.