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That is prove that the numerator of $\sum^{p-1}_{k=1} \frac{1}{k^2}$ expressed in lowest terms is divisible by $p$ where $p$ is prime.

I have tried to express the numerator but am not having much luck reducing it and I cannot find much online that is helpful here. Any hints/help is greatly appreciated, thanks in advance.

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  • $\begingroup$ GCD $ (a,b)=0?$ $\endgroup$ – lab bhattacharjee Oct 17 '17 at 4:14
  • $\begingroup$ @Labbhattacharjee fixed, thanks $\endgroup$ – Rick Owens Oct 17 '17 at 4:17
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Hint:

$$\sum_{k=1}^{p-1}\dfrac1{k^2}\equiv\sum_{r=1}^{p-1}r^2\pmod p$$

as $(k,p)=1$ for $1\le k<p,$ there exists a unique $r,1\le r<p$ such that $kr\equiv1\pmod p$

See also: Multiplicative Inverse of Modulo n is UNIQUE

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  • $\begingroup$ Because each element has a unique inverse mod p? I think I will be able to get it now, thanks! $\endgroup$ – Rick Owens Oct 17 '17 at 4:19
  • $\begingroup$ @Rick, Thanks for the rectification $\endgroup$ – lab bhattacharjee Oct 17 '17 at 4:25

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