# Stronger versions of Wilson's Theorem

## Problem

Let $$c \in \mathbb{N}$$ $$;$$ $$\exists$$ a prime $$p$$ for which:

$$p^c \mid (p-1)!+1$$

Does $$\exists$$ $$M$$ $$\in$$ $$\mathbb{N}$$ $$;$$ $$\forall$$ $$c \geqslant M$$ $$;$$ $$\nexists$$ $$p$$ satisfying the above?

## When $$c$$ = $$1$$

The statement is equivalent to Wilson's Theorem. For every prime $$p$$: $$p \mid (p-1)!+1$$ Proof:

$$\forall$$ $$x \in {1,2,...,p-1}$$ $$\exists!$$ $$x' \in {1,2,...,p-1}$$ ;

$$x \cdot x'\equiv 1 \pmod{p}$$

$$x=x' \iff p \mid x^2-1 \iff x = 1, x=p-1$$

$$(p-1)! = 1 \cdot (p-1) \cdot \prod{(x \cdot x')} \equiv 1^n \cdot (p-1) \equiv p-1 \pmod{p}$$

$$\implies p \mid (p-1)!+1$$

QED

## When $$c$$ = $$2$$

We have the statement: $$p^2 \mid (p-1)!+1$$ The only known primes that satisfy this are $$5$$, $$13$$ and $$563$$.

$$(5-1)!+1 = 25 = 5^2$$

$$(13-1)!+1 = 479001601 = 13^2 \cdot 2834329$$

$$563^2 \mid (563-1)!+1 \approx 1.128 \cdot 10^{1303}$$

Such primes $$p$$ are known as Wilson Primes. It is conjectured that there are infinitely many Wilson Primes. However, if there exists a fourth Wilson prime $$p$$, then $$p>2 \cdot 10^{13}$$.

## When $$c \geqslant 3$$

There are no known primes for which $$p^3 \mid (p-1)!+1$$ as if there is, then $$p$$ also has to be a Wilson Prime.

$$(5-1)!+1 = 25 \equiv 25 \pmod{5^3}$$

$$(13-1)!+1 = 479001601 \equiv 676 \pmod{13^3}$$

$$(563-1)!+1 \equiv 91921010 \pmod{563^3}$$

It is most likely due to following evidence that there exists an upper bound $$M$$ for which: $$c \geqslant M \implies p^c \nmid (p-1)!+1$$ where $$M \geqslant 3$$.

• We consider $$(p-1)!+1 \pmod{p^c}$$
• We assume that every remainder divisible by $$p$$ (Wilson's Theorem) is equally probable.
• Thus, the probability of required remainder $$0$$ is $$\frac{1}{p^{c-1}}$$
• Thus, probable number of primes for given constant $$c$$ is: $$\sum{\frac{1}{p^{c-1}}} = P(c-1)$$ where $$P(x)$$ is the Prime Zeta Function

When $$c=2$$, the expected number of Wilson primes is $$P(1)$$. $$P(1)=\sum{\frac{1}{p}}$$ This sum is divergent. Thus, it is probable that there exist infinitely many Wilson primes.

Proof:

Define $$N(x)$$ to be the number of positive integers $$n \leqslant x$$ for which $$p_i \nmid n$$, where $$i > j$$ for constant $$j$$ and $$p_i$$ is the $$i$$th smallest prime. Then, we write: $$n=k^2m$$ where $$m$$ is square-free.

As $$m$$ is square-free, and the only primes that divide it are $$p_i$$ for $$1 \leqslant i \leqslant j$$, it has $$2^j$$ possibilities.

$$n^2 \leqslant x \implies n \leqslant \sqrt{x}$$, thus giving $$n$$ a maximum of $$\sqrt{x}$$ possibilities.

$$\implies N(x) \leqslant 2^j\sqrt{x}$$

Assume the contrary, then for some $$j$$: $$\sum_{i=j+1}^\infty \frac{1}{p_i} < \frac{1}{2}$$ We also have $$x-N(x)$$ is the number of numbers less than or equal to $$x$$ divisible by one or more of $$p_i$$ for $$i>j$$. $$x-N(x) \leqslant \sum_{i=j+1}^\infty \frac{x}{p_i} < \frac{x}{2}$$ $$\implies 2^j\sqrt{x} > \frac{x}{2}$$ which is untrue for $$x \geqslant 2^{2j+2}$$

Thus the sum diverges. The divergence is similar to $$\log{\log{x}}$$ (Which is very slow).

When $$c \geqslant 3$$, the sum converges and is less than $$1$$.

When $$c=3$$, $$P(c-1) \approx 0.45$$

When $$c=4$$, $$P(c-1) \approx 0.17$$

When $$c=5$$, $$P(c-1) \approx 0.07$$

When $$c=6$$, $$P(c-1) \approx 0.03$$

When $$c=7$$, $$P(c-1) \approx 0.002$$

We now go on to show why there most probably exists a constant $$M$$ such as the one in the problem. Consider: $$\sum_{i=3}^\infty P(i-1)$$

We have:

$$\sum_{i=3}^\infty P(i-1) < \sum{\frac{1}{n(n+1)}} = \sum{\biggl(\frac{1}{n}-\frac{1}{n+1}\biggl)} = 1$$

Thus, the probable sum of the number of primes that satisfy the statement for $$c \geqslant 3$$, including a prime $$p$$, $$n-2$$ times, if the maximum $$c$$ satisfied is $$n$$, is less than $$1$$. However, if the answer to the problem is false, then, the sum would be infinite.

Thus, it is highly unlikely for there to be a solution for $$c \geqslant 3$$ as the probable answer is less than $$1$$ but the actual answer would be a positive integer. However, it is almost impossible for the answer to the problem to be false, as the probable answer is less than $$1$$ but the actual answer would be infinite!

Any of the following:

• Any progress or insight
• Polynomial or logarithmic non-trivial bounds on $$M$$
• Why are you adding over all $c \ge 3$??? Can't you just deduce the finiteness of the sum for $c=3$ and be done? – mathworker21 Sep 19 '18 at 17:49
• @DietrichBurde it is definitely an open problem since it is open for $c=2$ – mathworker21 Sep 19 '18 at 17:54
• Please read the actual problem. It is about whether there exists an upper bound $M$. Even if $M=10^{10^{10}}$, the problem is solved. One just needs to show its existence, if not even its value, however big. This is hence a weaker problem, which is especially portrayed in the last part by probability using prime zeta function. – Haran Sep 19 '18 at 17:57