Congruence for the sums of odd powers of integers Does someone know how to prove ***EDIT by induction**** that for all integers $n\ge1$, $k\gt0$  $$\sum\limits_{i=1}^{n} {i^{2k+1}}\equiv 0\  \ \  \pmod{\frac{n(n+1)}{2}}$$
I thought this should be a quick outcome from known polynomial expressions for sums of powers, or easy by induction. But I have not been able to write down such a proof.
Thanks for help..
I make the above EDIT to 'deduplicate' (somehow) the question.

This is where I am: 
Let $P_{m+1}$ be the $m+1$ degree polynomial such that $P_{m+1}(n) =\sum\limits_{i=1}^{n} {i^{m}}$
From 
$$(n+\frac{1}{2})^{m+1}-(\frac{1}{2})^{m+1}=\sum\limits_{i=1}^{n}((i+\frac{1}{2})^{m+1}-(i-\frac{1}{2})^{m+1})$$ after expansion of the binomes on the RHS and rearrangement:
$$(n+\frac{1}{2})^{m+1}-(\frac{1}{2})^{m+1}=\sum\limits_{j=0}^{m+1}\frac{\binom{m+1}{j}}{2^{j}}(1-(-1)^j)P_{m+2-j}(n)$$
I obtain a recurrence relationship for the $P_{2k}$ of even degrees
$$2^{2k+2}(2k+2)P_{2k+2}(n)=(2n+1)^{2k+2}-1-\sum\limits_{j=1}^{k}2^{2j}\binom{2k+2}{2j-1}P_{2j}(n)$$
that is 
$$2^{2k}(k+1)P_{2k+2}(n)=\binom{n+1}{2}\sum\limits_{j=0}^{k}(2n+1)^{2j}-\sum\limits_{j=1}^{k}2^{2j-3}\binom{2k+2}{2j-1}P_{2j}(n)$$
Now if I suppose that, for all $1\le j \le k$, for all $n$, all the $P_{2j}(n)\equiv 0\  \  \pmod{\frac{n(n+1)}{2}}$ , which is true for $j=1$, and since all the $2^{2j-3}\binom{2k+2}{2j-1}$ are integers even when $j=1$,  then I have
$$2^{2k}(k+1)P_{2k+2}(n)\equiv 0 \pmod{\frac{n(n+1)}{2}}$$
which is almost (but not quite :-() enough to complete the proof by induction 
 A: This is a work in progress. I'm not quite there yet, but it seems so close. I will start by deriving a more convenient recursion formula for $S_{2k+1} = \sum_{i=1}^n i^{2k+1}$. The formula follows very nicely from a combinatorial argument (due to Timothy Gowers, see his blog for all the details)

Lemma: Let $I_p = \sum_{i=0}^p S_{2k+1-i}(n)\left((-1)^i+1\right) {p\choose i}$ then$$\sum_{p=0}^k I_p = n^{k+1}(n+1)^{k+1}$$

The proof follows by computing the left hand side which reads
$$\sum_{r=1}^n\sum_{p=0}^k\sum_{i=0}^p r^{2k+1-i}\left((-1)^i+1\right) {p\choose i} = n^{k+1}(n+1)^{k+1}$$
We now do the $i$-sum using the binomial formula and then the $p$-sum which is just geometrical series to get
$$\sum_{r=1}^n r^{k+1}\left((r+1)^{k+1}-(r-1)^{k+1}\right)$$
which telescopes to give the desired sum $n^{k+1}(n+1)^{k+1}$.
Performing the sum in the lemma by grouping terms with the same $S_i(n)$ prefactor gives us
$$\sum_{i=0}^{\left\lfloor\frac{k}{2}\right\rfloor}{k+1\choose 2i+1}S_{2k+1-2i}(n) = \frac{n^{k+1}(n+1)^{k+1}}{2}$$
or
$$(k+1)S_{2k+1}(n) = \frac{n^{k+1}(n+1)^{k+1}}{2} - \sum_{i=1}^{\left\lfloor\frac{k}{2}\right\rfloor}{k+1\choose 2i+1}S_{2k+1-2i}(n)$$
Now assume for induction that $\frac{n(n+1)}{2}$ divides $S_{2r+1}$ for $r=1,2,\ldots,k-1$ then
$$(k+1)S_{2k+1}(n) \equiv 0 \mod \frac{n(n+1)}{2}$$
if $\gcd\left(k+1,\frac{n(n+1)}{2}\right) = 1$ then it follows that $\frac{n(n+1)}{2}$ divides $S_{2k+1}$ and the induction step is done. If this is not the case then its much harder to proceed and I have not yet found a way to continue...
However, we can state a partial result which follows from a simple induction argument: 

$S_{2k+1}(n)$ for $k\geq 1$ can be written as 
  $$S_{2k+1}(n) = n^2(n+1)^2 P_{2k-2}(n)$$
  where $P_i(n)$ is a polynomial of degree $i$ in $n$ with rational coefficients and leading term $\frac{n^{2k-2}}{2k+2}$.

