If $\sum_{k=1}^{p-1}\frac{1}{k^2}=\frac{b}{a}$, prove that $p\mid b$ 
Let $p>5$ be a prime. Suppose $$\sum_{k=1}^{p-1}\dfrac{1}{k^2}=\dfrac{b}{a},\quad a,b\in\mathbb{N}, (a,b)=1 $$ Prove that $p\mid b$.

Rewrite LHS$$\sum_{k=1}^{p-1}\dfrac{1}{k^2}=(\sum_{k=1}^{p-1}\dfrac{1}{k})^2-2\sum_{1\leq i<j\leq p-1}\dfrac{1}{ij}$$
Let $\sum_{k=1}^{p-1}\frac{1}{k}=\frac{d}{c}$, I have proven that $p^2|d$. However, I do not know how to handle the $\sum_{1\leq i<j\leq p-1}\frac{1}{ij}$ part.
Is my way ok and how to go on? If not, please suggest another method. Appreciate any help!
 A: The set $\{1,...,p-1\}$ under multiplication modulo $p$ is a (cyclic) group. Therefore for every $0<k<p$ there is unique $m$ between $1$   and $p-1$ such that $m\equiv 1/k\mod p$. Therefore
$$\sum 1/k^2\equiv \sum_1^{p-1} m^2 =\frac{p(p-1)(2p-1)}6\equiv 0\mod p$$ assuming $p>3$.
This implies the OP statement.
A: Consider the sum modulo $p$. In $\Bbb{Z} / p\Bbb{Z}$, we have
$$\sum_{k \in \Bbb{Z}_p^*} k^{-2} = \sum_{k \in \Bbb{Z}_p^*} k^2,$$
by simple variable substitution. Recall that,
$$\sum_{k=1}^{p-1} k^2 = \frac{1}{6}(p - 1)p(2p - 1),$$
hence if $p$ is a prime of at least $5$ (equal is allowed), then
$$\sum_{k \in \Bbb{Z}_p^*} k^{-2} = \sum_{k \in \Bbb{Z}_p^*} k^2 = 0.$$
Clearly the following is a natural number:
$$c := ((p - 1)!)^2\sum_{k=1}^{p-1} \frac{1}{k^2},$$
but modulo $p$, this is $0$ by our previous arguments. Thus, we may express
$$\sum_{k=1}^{p-1} \frac{1}{k^2} = \frac{c}{(p - 1)!)^2},$$
where $p \mid c$. Note also that $p \not\mid ((p - 1)!)^2$. By reducing the fraction on the right hand side, we therefore cannot remove the factor of $p$ from $c$, hence we must have $p \mid a$.
