I need to prove this.
Let $p> 2$ be an odd prime and let $n$ be a positive integer. Prove that
$p$ divides $\sum\limits_{k=1}^{p-1} (k^p)^n $
Hint: Be $k=p^n$ and evaluate $d^k+(p-d)^k$ use the binomial theorem.
Thanks you for help me!
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Change title to describe content of question. – mlc Mar 04 '17 at 21:37
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1What's $d{}{}$? – Mar 04 '17 at 21:40
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1Use the hint you are given, use the binomial theorem to write out $(p-d)^k$ , then work out divisibility.. – victoria Mar 04 '17 at 21:41
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It’s not true, as you see from the example $p=3$, $n=2$. In general, if $n$ is a multiple of $p-1$, you’re taking (modulo $p$) the sum of $1$ just $p-1$ times, giving a sum that’s $\equiv-1\pmod p$.
Lubin
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Because $\gcd(p,k)=1, \forall k=\overline{1,p-1}$, from Fermat's little theorem $$k^p\equiv k \pmod{p} \Rightarrow (k^p)^n\equiv k^n \pmod{p}$$ then $$\sum_{k=1}^{p-1}(k^p)^n\equiv \sum_{k=1}^{p-1}k^n \pmod{p}$$ As Lubin showed, the statement isn't always true. However, it works for $n=1$ because $$\sum_{k=1}^{p-1}k=\frac{p(p-1)}{2} \equiv 0 \pmod{p}$$ It works for $n=3$, because $$\sum_{k=1}^{p-1}k^3=\left(\frac{p(p-1)}{2}\right)^2 \equiv 0 \pmod{p}$$ And generally, for any $n=2q+1$, i.e. odd, because, due to binomial expantion $$\sum_{k=1}^{p-1}k^{2q+1}=(1+(p-1)^{2q+1}) + (2^{2q+1}+(p-2)^{2q+1})+ (3^{2q+1}+(p-3)^{2q+1})+...+\left(\left(\frac{p-1}{2}\right)^{2q+1}+\left(p-\frac{p-1}{2}\right)^{2q+1}\right) \equiv 0 \pmod{p}$$
rtybase
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@BellaPatarroyo it's not too difficult, $K \equiv 0 \pmod{p}$ is similar to $p | K$ or $p$ divides $K$. – rtybase Mar 05 '17 at 00:42
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