What is the value of ? ? For any positive integer k, find the value of and prove your assertion is correct?
So I simply tested out values, and for a lot of primes and powers, it's simply 0, but sometimes, when , for example, it's sometimes either 0 or 2, like when it's 2, but when , it's 0. I can't come up with a clear formula for any of these values.
Isn't it actually either or depending on whether k is odd or even? I mean, I understand that you'd use the binomial theorem and that all terms except the term has factors of p in it, and are therefore 0 modulo p.
Well, this is for number theory class so we haven't covered or used anything regarding rings so we really couldn't use anything like that on an exam. And yeah, it is easy to show when just k=1 that it's 0 for all primes greater than 2, but when it gets to squares or k in general, I'm still lost and I can't figure out a general expression of the value.
I mean, if k is even then the summation of the k-th powers of 1 through p-1 modulo p is Other than knowing that, I'm lost.
My point was that if then the power sum formula you know and love =\frac{1}{2}n(n+1)[/tex] is true, if then holds true, etc. Namely, if you have a power sum formula and you're dealing with a prime which doesn't divide the denominator of the formula then it still works.
Formula for what? If you mean modulo the prime, then as Unbeatable pointed out it's zero if the prime is odd and one otherwise (this is clear from what I said as well since always divides the sum). If you mean for the actual formulas, there is no nice formula. Did you look at that link that I sent you?
it seems to me that by post #2 (and #5), we know that .
if p = 2, we always get 1, no matter what k is. otherwise p is an odd prime.
if k is odd, we can pair n with p-n, we have (p-1)/2 such pairs, which will always sum to 0 (mod p).
if k is even, the situation is more complicated. for example:
but we cannot apply this formula for p = 2,3 or 7. indeed,
"most" of the sums of 6th powers will sum to 0 (mod p), but we will have exceptions (prime divisors of 42).
i suspect that, in the even k case, the exceptions always sum to p-1 (mod p), but i have not seen (in this thread, at any rate) a proof of that, or a way to enumerate which p will prove exceptional.
this is untrue.
What is untrue about it? Perhaps I was unclear in my wording. It's zero when the prime is odd for the case . Beyond that it's not clear how exactly it reacts. It's clear though, as I was trying to get across, that when summing gives you zero is highly connected to the constant in the denonminator of the formula for the sum since, if a prime doesn't divide this denominator then the sum is zero.