When is $sum_{i=1}^{m}i^n =0 bmod dfrac{m(m+1)}{2}$?

Mathematics Asked on January 5, 2022

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 ?

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