Thread: p|((n^p)-n)

Someone ask me to prove that p|((n^p)-n) for p=prime numbers and n=all integers

I don’t know how to write a formal proof. But I know that the problem has 2 condition :

Condition 1
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)

Condition 2

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

For example
3|((5^3)-5)

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 ??
thanx...

It is a wonderful and important theorem first proposed by my favorite mathemation, Fermat.

Klicken Heir.

Wow, look how many proofs there are. Even through Dynamical systems, who could have imagned. If thou with I can post my own proof of the little theorem.

Originally Posted by ThePerfectHacker

.... I can post my own proof of the little theorem.

please ....

thanks anyway