Let $p\ge5$ be a prime. If the following fraction is fully reduced (to an irreducible fraction), prove that the numerator will be divisible by $p^4$.
$\sum_{k=1}^{p^2} \frac1k - \frac1p\sum_{k=1}^p\frac1k$
I have a solution to this problem, but I can't understand some parts of it. (Though it is written in Korean, you will have no difficulty in understanding it.)
First of all, I just know about modular arithmetic of integers, but what means that a quotient is congruent to 0 ($\dfrac ab \equiv 0 \mod m$)? (In this problem, $\displaystyle \sum_{i=1,p\nmid i}^{(p^2-1)/2}\frac1{i(p^2-i)} \equiv 0\mod p^2$)
Second, why $\displaystyle \sum_{i=1,p\nmid i}^{p^2}\dfrac1{i^2} \equiv \sum_{i=1,p\nmid i}^{p^2}i^2 \mod p^2$ does hold?