Dtyjlui

Let $pgt 3$ 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.

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 $(p-1)!$].

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

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

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

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