Congruence for the sums of odd powers of integers [duplicate]

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

marked as duplicate by TonyK, Jyrki Lahtonen, Rick Decker, Micah, hardmathNov 17 '14 at 0:34

• math.stackexchange.com/questions/427744/… – lab bhattacharjee Nov 16 '14 at 18:24
• @lab bhattacharjee, thank you so much! I was suspecting that there was a trick to obtain a direct proof without need of these complicated recurrences and induction (which by the way don't do the job, so far), but I could not find it. – René Gy Nov 16 '14 at 18:59

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 $\frac{n^{2k-2}}{2k+2}$.

• Thank you! That is enlightening. Your recurrence relation for the sums of odd powers is better than mine because you get rid of my $2^{2k}$ factor. But isn'nt there still something missing? If your induction hypothesis is that $S_{2m+1}(n) \equiv 0 \pmod{\frac{n(n+1)}{2}}$ for all $1\le m \lt k$, you then obtain $$(k+1)S_{2k+1}(n) \equiv 0 \pmod{\frac{n(n+1)}{2}}$$ In order to conclude that $$S_{2k+1}(n) \equiv 0 \pmod{\frac{n(n+1)}{2}}$$ don’t we need that $k+1$ and $S_{2k+1}(n)$ be coprime for all $k$ and all $n$ ? – René Gy Nov 16 '14 at 9:42
• @RenéGy You were right about that. Can't see right now how I should close the gap, but I haven't given up yet. – Winther Nov 16 '14 at 21:07
• Would it help, if you replace $\sum_{p=2j+2}^k {p\choose 2j+2}$ by its value ${k+1\choose 2j+3}$ ? – René Gy Nov 16 '14 at 21:45