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### Fermat's Theorem (Not the Last)

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Date: 07/14/99 at 13:07:44
From: Sandy Nicholson
Subject: Fermat's theorem (not the last)

Reading "A Beautiful Mind," about John F. Nash, pg 35 mentions that at
age 14 Nash proved Fermat's theorem: If n is a whole number and p any
prime, then n multiplied by itself p times minus n is divisible by p.
I can't prove it and am looking for proof.

Thanks by the way for Dr. Math; I use it and have grown from it.
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Date: 07/14/99 at 15:26:38
From: Doctor Anthony
Subject: Re: Fermat's theorem (not the last)

If p is a prime number and N is prime to p then

N^(p-1) - 1   is a multiple of p.  [= M(p)]

If you have an expression like (a+b)^2 = a^2 + b^2 + 2ab

(a+b)^n = a^n + b^n +  M(n)

(a+b+c)^n = a^n + b^n + c^n + M(n)  and so on

(a+b+c+......)^p  = a^p + b^p + c^p + ..... + M(p)

Now put a=b=c=...... = 1 and suppose there are N such letters

(1+1+1+... to N terms)^p = 1^p + 1^p + .....+ 1^p + M(p)

N^p = N + M(p)

N^p - N = M(p)

N[N^(p-1) - 1] = M(p)

but N is prime to p so we MUST have

N^(p-1) - 1 = M(p)

This is the actual statement of Fermat's theorem.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
College Number Theory

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