Prove that numerator of the sum $\sum_{n=1}^{p-1}\frac{1}{n^2}$ is divisible by $p$ if $p$ is prime.

Question

Assume $p$ is a prime number such that $p>3$ and $\gcd(n,m)=1.$

If: $$1 +\frac{1}{2^2} +\frac{1}{3^2} +\cdots+\frac{1}{(p-1)^2} =\frac{m}{n}$$

for some integers $m,n$, what is the proof that $p\mid m?$

Answer 1

Start with the given equality in $\Bbb Q$. Then the denominator $n$ is a divisor of $((p-1)!)^2$ (edit, thanks user!), so is prime to $p$. Multiply with $n$. This gives an identity in $\Bbb Z$. We consider it modulo $p$, and work from now on only in $$ F = \Bbb Z/p = \Bbb F_p\ ,$$ in the field with $p$ elements. We have to show the corresponding sum is zero in $F$. Now we divide by $n$ (taken modulo $p$), have to show equivalently in $F$: $$1 ^{-2}+2^{-2}+\dots+(p-1)^{-2}=0\text{ in }F=\Bbb F_p\ .$$ The "numbers" in $F$ obtained by taking the multiplicative inverses, $1/1$, $1/2$, ... $1/(p-1)$ is in an other order the same set $1,2,\dots,(p-1)$. So we have to show $$1 ^{2}+2^{2}+\dots+(p-1)^{2}=0\text{ in }F=\Bbb F_p\ .$$ For the L.H.S. we have a formula in $\Bbb Z$, which involves the denominator $6$, without problems in our case, $p>3$, and which comes with the factor $((p-1)+1)\equiv 0$ modulo $p$.

$\blacksquare$