Let $p$ be an odd prime and consider the sum $$1+\frac{1}{2}+\cdots+\frac{1}{p-1}=\frac{a}{b},$$ where $a,b$ are integers. Then I showed that $p \mid a$.
But now I have to show that if $p>3$, a prime, and if $$1+\frac{1}{2}+\cdots+\frac{1}{p-1}=\frac{a}{b},$$ where $a,b$ are integers, then $p^2 \mid a$.
For the first part I did the following. Consider the group homomorphism $\Bbb{F}_p^* \to \Bbb{F}_p^*$ given by $x\mapsto x^{-1}$. Since this map is one to one this is a bijection, $\Bbb{F}_p^*$ being a finite set. Thus $$1+\frac{1}{2}+\cdots+\frac{1}{p-1}=1+2+\cdots+(p-1)=\frac{p(p-1)}{2}$$, thus $pb(p-1)=2a\implies p \mid a$, since $p$ is an odd prime. But this process no longer works for the next part. What can I do? Any help will be appreciated. Thanks.