$\sum \limits_{k=1}^{p-1}\frac{(p-1)!}{k} \equiv 0 \ [p]$?

Question

Here's the expression :

$\frac{(p-1)!}{1}+\frac{(p-1)!}{2}...+\frac{(p-1)!}{p-1}= (p-1)\times...\times2+(p-1)\times...\times1+...+(p-2)\times...\times1=q$.

For $p=5$ we obtain : $(4.3.2)+(4.3.1)+(4.2.1)+(3.2.1)=50$

I precise that $q\in \mathbb{N}^*$, and $p\ge 3$ a prime number.

We know that there are $(p-2)$ terms of each $k\in\{1,...,p-1\}$

$p$ divides $q$ ?

I don't search for a polynomial method or an arrangement of the coefficients. Just by developing is it possible ?

Thanks in advance !

Answer 1

Method 1

For distinct $k, \ell \in \{ 1, 2, \ldots, p-1 \}$, we have

$$\frac{(p-1)!}{k} - \frac{(p-1)!}{\ell} = \frac{(p-1)!}{k\ell}(k - \ell)\tag{*1}$$ The first factor is a product of numbers relative prime to $p$. The second factor $k - \ell$ is a number with $0 < |k - \ell| < p$. So RHS and hence LHS of $(*1)$ is relative prime to $p$. This means

$$\frac{(p-1)!}{k} \not\equiv \frac{(p-1)!}{\ell} \pmod p \quad\text{ for distinct }\quad k, \ell \in \{ 1,2,\ldots,p-1 \}$$

So under modulus $p$ arithmetic, $\frac{(p-1)!}{k}$ is simply a relabeling of the $p-1$ equivalent classes of integers relative prime to $p$. As a result,

$$\sum_{k=1}^{p-1}\frac{(p-1)!}{k} \equiv \sum_{\ell=1}^{p-1} \ell = \frac{p(p-1)}{2} \equiv 0 \pmod p$$

Method 2 $$\sum_{k=1}^{p-1}\frac{(p-1)!}{k} = \sum_{k=1}^{\frac{p-1}{2}}\left[\frac{(p-1)!}{k} + \frac{(p-1)!}{p-k}\right] = \sum_{k=1}^{\frac{p-1}{2}}\frac{(p-1)!}{k(p-k)}((p-k) + k)\\ = \sum_{k=1}^{\frac{p-1}{2}}\frac{p!}{k(p-k)} = p \left[ \sum_{k=1}^{\frac{p-1}{2}}\frac{(p-1)!}{k(p-k)}\right]\tag{*2} $$

Notice for each $1 \le k \le \frac{p-1}{2}$, $\displaystyle\;\frac{(p-1)!}{k(p-k)} = \prod_{\ell = 1, \ne k, p-k}^{p-1} \ell$ is a product of integer, so it is an integer.
This implies the sum in the square bracket on RHS of $(*2)$ is an integer. As a result, RHS and hence LHS of $(*2)$ is a multiple of $p$.

Answer 2

Apply Wilson's Theorem: $$(p-1)!\equiv -1 \pmod p\implies \frac {(p-1)!}k\equiv -k^{-1} \pmod p$$ For any non-zero residue $p$.

Thus the desired sum is just $$-\sum_{l=1}^{p-1}k^{-1}$$ But, up to sign, that's just the sum of the residues (as no two residues have the same inverse). As $p$ is odd, the sum of the residues $\pmod p$ is $0$.

As a side note, OEIS A007619 gives the sequence of "Wilson's quotients", i.e. $\frac {(p-1)!+1}p$ for primes $p$. There is no reference there to any alternate expression for them. Doesn't prove there isn't a simple expression which gives them, of course, nor does it prove that there isn't a simple expression for your quotient.