show that 1^p+2^p+3^p+...+(p-1)^p is congruent to 0 (mod p) when p is an odd prime
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It can be proved for any odd integer...
Separate the sum in two parts :
In each sum, there are terms.
And in the second one, just notice that :
Or a bit more simply (no offense Moo ) note that
by Fermat's little theorem. Hence
Originally Posted by Bruno J. Or a bit more simply (no offense Moo ) note that
by Fermat's little theorem. Hence No offence
It's just that I've taken the proof from the case p is an odd integer, not necessarily a prime number
Well, surely yours is easier and better for this case
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