Let $p\gt 3$ be an prime. If $\sum_{k=1}^{p-1} \frac{1}{k}=\frac{a}{b}$, where $\gcd(a,b)=1$. Prove that $p\mid a$.

Question

Let $p\gt 3$ be an prime. Suppose $$\sum_{k=1}^{p-1} \frac{1}{k}=\frac{a}{b}$$ where $gcd(a,b)=1$. Prove that $a$ is divisible by $p$.

Please give me some hint. Sorry for this types of writing. I am not familiar with this.

Answer 1

Well

$$\sum_{k=1}^{p-1} \frac{1}{k} \ = \ \sum_{k=1}^{p-1} \frac{\frac{(p-1)!}{k}}{(p-1)!}$$

Now $\frac{(p-1)!}{k}$ is an integer for each such $k$, so writing $A=(p-1)!$, we note that $\frac{A}{k} \equiv_p (A \mod p)(k^{-1} \mod p)$. Thus

$$\sum_{k=1}^{p-1} \frac{(p-1)!}{k} \ \doteq \ \sum_{k=1}^{p-1} \frac{A}{k} \ \equiv_p \ \sum_{k=1}^{p-1} (A \mod p)(k^{-1} \mod p).$$

WE then use the fact that each element $k$ has a unqiue inverse to conclude

$$\sum_{k=1}^{p-1} (A \mod p)(k^{-1} \mod p) \ \equiv_p \ \sum_{k=1}^{p-1} (A \mod p)k.$$

However, one can check that for a prime $p \geq 3$ the following holds: $\sum_{k=1}^{p-1} (k \mod p) \equiv_p 0$, concluding

$$\sum_{k=1}^{p-1} (A \mod p)(k^{-1} \mod p) \ \equiv_p \ \sum_{k=1}^{p-1} (A \mod p)k \ \equiv_p \ 0,$$

which gives you what you need to show [make sure you see why, it follows because $p$ does not divide $(p-1)!$].

Answer 2

For $k\leq {p-1\over 2}$ let $q_k := {1\over k(p-k)}$

So we have $$({1\over 1}+{1\over p-1})+({1\over 2}+{1\over p-2})+...+({1\over {p-1\over 2}}+{1\over {p+1\over 2}}) = {a\over b}$$

$$pq_1+pq_2+...+pq_{p-1\over 2} = {a\over b}$$

Let $$q_1+q_2+...+q_{p-1\over 2} = {c\over (p-1)!}$$ for some integer $c$. So we have $$p\cdot c \cdot b = a\cdot (p-1)!\implies p\mid a\cdot (p-1)!\implies p\mid a$$