# Show that the numerator of $\sum_{k=1}^{p^2} \frac1k - \frac1p\sum_{k=1}^p\frac1k$ is divisible by $p^4$

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?

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$a/b\equiv c/d\pmod m$ simply means $ad\equiv bc\pmod m$. In particular, $a/b\equiv0\pmod m$ just means $a$ is a multiple of $m$ (assuming $a/b$ is in lowest terms).
If $p$ doesn't divide $i$ then there is $j$ such that $ij\equiv1\pmod p$ so $1/i^2\equiv j^2$, and as $i$ runs through the residues mod $p^2$, so does $j$.