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proof the integer: From Fermat's little theorem, if $\gcd(a,p) = 1$, then we have: $a^p \equiv a \pmod p$. Since $p$ is only divisible by itself and 1, so we have: $1^p+2^p+3^p+...+(p-1)^p \equiv 1+2 + ...+ p-1 \pmod p$

$\equiv p(p-1)/2$. Since p is odd prime, $p -1 = 2k$. So we have:

$p(p-1) = kp\equiv 0 \pmod p$. Therefore $L_p$ is an integer.

The question is how can I generalize the integer $L_p$ for all odd integers $p$. That is, in the expression for $L_p$ replace the exponent $p$ with another non-constant quantity depend on $p$ in a way that $L_p$ remains an integer for all odd integer $p$, and justify the choice of exponent.

Martin Sleziak
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user3591471
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1 Answers1

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For any odd number $p$, $p$ divides $k^p+(p-k)^p$ for $1\leq k\leq p-1$: $$k^p+(p-k)^p=k^p+\sum_{j=0}^p\binom{p}{j}p^{p-j}(-k)^j=\sum_{j=0}^{p-1}\binom{p}{j}p^{p-j}(-k)^j=pm.$$ where $m:=\sum_{j=0}^{p-1}\binom{p}{j}p^{p-1-j}(-k)^j$ is an integer. Then it follows that $p$ divides also $$\sum_{k=1}^{p-1}k^p=\sum_{k=1}^{(p-1)/2}(k^p+(p-k)^p).$$

Robert Z
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