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I’m looking at the sum of m-th powers of consecutive integers

i.e. $Sm(p) = 1^m + 2^m +…+(p-1)^m$

I need to prove: Let $p$ be an odd prime. Prove the following congruences:

$$\begin{align} Sm(p^2) &\equiv 0\pmod{p^2}\text{ if $(p − 1)$ does not divide m}\\ Sm(p^2) &\equiv -p\pmod{p^2}\text{ if $(p − 1)$ divides m} \end{align}$$

For all $m > 1$

I recognize that in order to solve using primitive roots I will need to eliminate the multiples of p in the expansion.

Could someone please show me how to prove it?

Thomas Andrews
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user319032
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1 Answers1

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Partial Solution.

If we assume that $m$ is odd, then we know that $(p-1)$ does not divide $m$ (as $p-1$ is even). We are then in the first case. In this situation, the claim follows from the fact that $$m\;odd\implies\frac {(n-1)n}2\;|\;S_m(n)$$

(see, eg, here )

lulu
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