Proof that $\sum\limits_i \frac{(p-1)!}i$ is divisible by $p$

Solution 1:

This is not a complete proof but rather some thoughts.

Edit: I erroneously states that this works for p=2 when it does not.

Suppose $p$ is an odd prime. By the hint, we have that $$p^{p-1} - 1 \equiv (p-1)! \pmod p.$$ On the other hand $$p^{p-1} -1 \equiv -1 \pmod p.$$ Putting these two facts together gives $$(p-1)! \equiv -1 \pmod p.$$ Therefore, $$K \equiv \sum_i^{p-1}\frac{-1}{i} \pmod p.$$ Can you finish from here?

Solution 2:

I have a much simpler solution. But I will start by stating that this is problem number 60 from this book (more complicated since it asks for $p^2 \mid K, p>3$) and the solution below is not from the book. So, for $p>2$ we have $p-1$ even and $$\sum\limits_{i=1}^{p-1}\frac{1}{i}=\sum\limits_{i=1}^{\frac{p-1}{2}}\frac{1}{i}+\sum\limits_{i=\frac{p+1}{2}}^{p-1}\frac{1}{i}= \sum\limits_{i=1}^{\frac{p-1}{2}}\frac{1}{i}+\sum\limits_{i=1}^{\frac{p-1}{2}}\frac{1}{p-i}=\\ \sum\limits_{i=1}^{\frac{p-1}{2}}\left(\frac{1}{i}+\frac{1}{p-i}\right)= \sum\limits_{i=1}^{\frac{p-1}{2}}\frac{p}{i(p-i)}=p\sum\limits_{i=1}^{\frac{p-1}{2}}\frac{1}{i(p-i)}=p\frac{A}{(p-1)!}$$ where $K=pA$ and $A \in \mathbb{N}$.