Prove that $a\equiv 0 $ $(mod$ $p)$ given $\sum^{p-1}_{k=1} \frac{1}{k^2} = \frac{a}{b}$ where $p$ is prime and $gcd(a,b)=1$

Question

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.

@Labbhattacharjee fixed, thanks – Rick Owens Oct 17 '17 at 04:17 — Rick Owens, Oct 17 '17 at 04:17

lab bhattacharjee · Accepted Answer · 2017-10-17T04:26:19.077

2

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

edited Oct 17 '17 at 04:26

answered Oct 17 '17 at 04:16

lab bhattacharjee

274,582

Because each element has a unique inverse mod p? I think I will be able to get it now, thanks! – Rick Owens Oct 17 '17 at 04:19
@Rick, Thanks for the rectification – lab bhattacharjee Oct 17 '17 at 04:25

Prove that $a\equiv 0 $ $(mod$ $p)$ given $\sum^{p-1}_{k=1} \frac{1}{k^2} = \frac{a}{b}$ where $p$ is prime and $gcd(a,b)=1$

1 Answers1