**Definitions:**

Lagrange's theorem implies that for each prime $p$, with the exception of $±1$, the factors of $(p − 1)!$ can be arranged in unequal pairs, where the product of each pair $≡ 1 \pmod p$. *[See Wiki article on Wilson's theorem.](https://en.wikipedia.org/wiki/Wilson's_theorem#Prime_modulus)*

From the example in the link above, for $p=11$ we have

$$(11-1)!=[(1\cdot10)]\cdot[(2\cdot6)(3\cdot4)(5\cdot9)(7\cdot8)]  \equiv [-1]\cdot[1\cdot1\cdot1\cdot1]  \equiv -1 \pmod{11}$$

Let the products of the pairs that $≡ 1 \pmod p$ be the set $A$. For the above example then, $A=\{(2\cdot6),(3\cdot4),(5\cdot9),(7\cdot8)\}=\{12,12,45,56\}$.

**Conjecture:**

$$\lim_{n\rightarrow\infty}\dfrac{1}{p_n}\sqrt{\sum_{k \in A}\dfrac{k-1}{p_n}}\approx\dfrac{1}{\sqrt{8}}$$

where $p_n$ is the $n$th prime.

**Examples:**

For $p=11$ we have

$$\dfrac{1}{11}\sqrt{\dfrac{(11+11+44+55)}{11}}=\dfrac{\sqrt{11}}{11}=0.30151\dots$$

For $p=997$ we have

$$\dfrac{\sqrt{123589}}{997}=0.35355\dots$$

**Comments:**

I have no idea whether the above statement is correct, or how to go about trying to find a proof. Any comments on the any of the above are most welcome.