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I have tried to creat such formula related to divisibility using triangulair numbers and series , I have got the following problem such that I ask when is $S_n(m)=\sum_{i=1}^{m}i^n =0 \bmod \dfrac{m(m+1)}{2}$ ? , My simple attempt without using any complicated solution, I got that $n$ must be odd positive and integer and $m$ must be arbitrary positive integer , such that I treated only the first case when both $n$ and $m$ are odd we may group the terms of $ S_n(m)$ as follows, and as $n$ is also odd we see by expanding the binomial that : $S_n(m)=m^n+\sum_{i=1}^{(m-1)/2}(i^n+(m-i)^{n})$ this means only that $m | S_n(m)$ but How I can follow this idea to pove that: $m(m+1)/2 | S_n(m)$ if what i have claimed is true ?

zeraoulia rafik
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    https://math.stackexchange.com/questions/427744/showing-that-1k2k-dots-nk-is-divisible-by-nn1-over-2 – lab bhattacharjee Jul 24 '20 at 23:32
  • Thanks for the link – zeraoulia rafik Jul 24 '20 at 23:40
  • @zeraouliarafik FYI, on a closely related note, you may be interested in reading about Faulhaber's formula. As you can see in that Wikipedia article, with $a = \frac{n(n+1)}{2}$, the sums of the odd powers can be defined as polynomials, possibly divided by a constant integer, of $a$, with all of the sums of powers $\gt 1$ actually having a factor of $a^2$ (apart from possibly cases where the denominator, e.g., $3$ or $5$, is a factor of $a$). – John Omielan Jul 25 '20 at 01:54

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