Fermat's Theorem (Not the Last)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. 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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