Divisibility of a summation by $p^2$ I try to use the hint of this problem but I could not. Any detailed answer will be appreciated!
Let $p$ be a prime number which $p>3$, and $$a/b:=1+1/2+1/3+\cdots +1/(p-1).$$  How could we show that  $p^2 | a$?
(Hint: Consider the polynomial $f(x)=(x-1)(x-2)\dots (x-(p-1))$ and  study the coefficient of $x$ when $x=p$).
 A: Since $p\gt3$ is odd, it's convenient to write the polynomial $f(x)$ (which has an even number of roots) as
$$f(x)=a_0-a_1x+a_2x^2+\cdots+x^{p-1}$$
where, by Vieta's formulae, we have
$$a_0=1\cdot2\cdots(p-1)=(p-1)!\quad\text{and}\quad a_1=\left({1\over1}+{1\over2}+\cdots+{1\over p-1} \right)a_0={a\over b}(p-1)!$$
Note that $b\mid(p-1)!$, which is not divisible by $p$, so $a_1=ak$ for some integer $k\not\equiv0$ mod $p$.
We also have $f(p)=(p-1)(p-2)\cdot((p-(p-1))=(p-1)!$  Putting all this together, we see that
$$(p-1)!=f(p)=(p-1)!-akp+a_2p^2+\cdots\equiv(p-1)!-akp+a_2p^2\mod p^3$$
or
$$ak\equiv a_2p\mod p^2$$
To wrap things up, we need to show $a_2\equiv0$ mod $p$ if $p\gt3$ (and we should also see what doesn't work when $p=3$).  But working mod $p$, we have
$$\begin{align}
f(x)&=(x-1)(x-2)\cdots(x-{p-1\over2})(x-{p+1\over2})\cdots(x-(p-1))\\
&\equiv(x-1)(x-2)\cdots(x-{p-1\over2})(x+{p-1\over2})\cdots(x+1)\\
&=(x^2-1)(x^2-4)\cdots(x^2-({p-1\over2})^2)
\end{align}$$
and thus
$$a_2\equiv(-1)^{(p+1)/2}(1^2+2^2+\cdots+({p-1\over2})^2)$$
Recalling the general formula $1^2+2^2+\cdots+n^2={n(n+1)(2n+1)\over6}={n+1\choose2}{2n+1\over3}$, we find, by setting $n={p-1\over2}$, so that $2n+1=p$,
$$\pm3a_2\equiv {n+1\choose2}p\equiv0\mod p$$
This is where we need to assume $p\gt3$ in order to conclude $a_2\equiv0$ mod $p$.
