Re: Summation up to p-1 modulo p, p prime

the original question asked for what is: $\displaystyle \sum_{n=1}^{p-1} n^k\ (mod\ p)$? perhaps the difficuly is in what you meant by "it" in your post...your "it" and Pinkk's "it" may be 2 different things.

your orignal post (#11) was clearer.

i think this is an interesting problem, which "as-stated" doesn't have a definitive answer. i haven't looked very far into it, but my feeling is, that if p is a divisor of the "denominator" for a closed form expression of $\displaystyle \sum_{n=1}^{p-1} n^k$, each power will resolve to 1, and we get p-1. the "denominators" in question are the even-numbered terms in this sequence:

A064538 - OEIS

which are also the denominators of the even Bournoulli numbers, expressed in "lowest terms". it seems to me that this question is a "hard" question, which is belied by its simple statement.

the complete answer of this question, then, rests on proving the following:

let k be n even integer, and let $\displaystyle B_k$ denote the k-th Bernoulli number. if p divides the denominator of $\displaystyle B_k$ written in lowest terms, then $\displaystyle \sum_{n=1}^{p-1} n^k = p-1\ (mod\ p)$

conjecture: the statement "p divides the denominator..." can be replaced by p-1 divides k.

any takers?

Re: Summation up to p-1 modulo p, p prime

From my previous post here we know that $\displaystyle S_p^n \equiv 0 \pmod{p}$ when $\displaystyle 0\le n \le p-2$.

If $\displaystyle n=k(p-1)$ for some $\displaystyle k\in\mathbb{N}$ then $\displaystyle S_p^n \equiv p-1 \pmod{p}$ by Fermat's Little Theorem. Otherwise, for $\displaystyle n \ge p-2$ which is not divisible by $\displaystyle p-1$ we get $\displaystyle S_p^n \equiv S_p^r \pmod{p}$ where $\displaystyle 1 \le r \le p-2$ is the remainder of $\displaystyle n$ upon division by $\displaystyle p-1$. So again we obtain $\displaystyle S_p^n \equiv 0 \pmod{p}$. In conclusion:

For $\displaystyle n=k(p-1)$, $\displaystyle k\in\mathbb{N}$ we have $\displaystyle S_p^n \equiv p-1 \pmod{p}$, and otherwise $\displaystyle S_p^n \equiv 0 \pmod{p}$.

Re: Summation up to p-1 modulo p, p prime

Quote:

Originally Posted by

**Unbeatable0** From my previous post here we know that $\displaystyle S_p^n \equiv 0 \pmod{p}$ when $\displaystyle 0\le n \le p-2$.

If $\displaystyle n=k(p-1)$ for some $\displaystyle k\in\mathbb{N}$ then $\displaystyle S_p^n \equiv p-1 \pmod{p}$ by Fermat's Little Theorem. Otherwise, for $\displaystyle n \ge p-2$ which is not divisible by $\displaystyle p-1$ we get $\displaystyle S_p^n \equiv S_p^r \pmod{p}$ where $\displaystyle 1 \le r \le p-2$ is the remainder of $\displaystyle n$ upon division by $\displaystyle p-1$. So again we obtain $\displaystyle S_p^n \equiv 0 \pmod{p}$. In conclusion:

For $\displaystyle n=k(p-1)$, $\displaystyle k\in\mathbb{N}$ we have $\displaystyle S_p^n \equiv p-1 \pmod{p}$, and otherwise $\displaystyle S_p^n \equiv 0 \pmod{p}$.

excellent! right, by FLT, each summand is congruent to 1 mod (p-1) (because p-1 is the order of the multiplicative group, and p-1 divides the power we're raising to). i should have thought of that myself. and every other case, we can reduce mod p to knock the exponents down to less than p-1.