Someone ask me to prove that p|((n^p)-n) for p=prime numbers and n=all integers
I donít know how to write a formal proof. But I know that the problem has 2 condition :
p|n^p and p|n so p|((n^p)-n)
(if d|a and d|b then d|(a-b)
For example 2|((4^2) - 4)
1.(n^p)/p = k + r where k are integers or zero and r are rational numbers
2.n/p = m + r where m = zero or integers and r are rational numbers
1.(5^3)/3= 41 + (2/3)
2.5/3 = 1+ 2/3
1 and 2 produce the same r so ((n^p)-n)/p = (n^p)/p + n/p = k+r - (m+r) = k - m + r - r = k -m
since k and m is also integer or zero so (k - m) is also integer or zero then for condition 2 it's true that p|((n^p)-n)
Since for all n = integers and p= primes always fullfiled this two condition I assume that p|((n^p)-n) is true for all p=primes and n=integers
Is it right or there is a better way to prove it ??
February 26th 2007, 07:37 PM
It is a wonderful and important theorem first proposed by my favorite mathemation, Fermat.