New conjecture? $(\varphi(n))! = -1 \pmod n \iff n$ is prime (nearly the same as Wilson's)

Question

New conjecture? $(\varphi(n))! = -1 \pmod n \iff n$ is prime (nearly the same as Wilson's)

143 Views Asked by Bumbble Comm At 27 Mar 2026 - 12:06

$$ (n-1)! = -1 \pmod n \text{ iff } n \text{ is prime}, \text{ is Wilson's theorem,} $$

But coincidentally for now the expression passed to factorial is $n - 1$ which is (iff $n$ is prime) equal to $\varphi(n)$. So, I was thinking, don't we have this related conjecture:

$$ \varphi(n)! = -1 \pmod n \text{ iff } n \text{ is prime} $$

Because the following example verification code seems to indicate as such:

from sympy import *

N = 10000

for n in range(2, N):    
    r = factorial(totient(n)) % n
    
    if isprime(n):
        assert (r + 1) % n == 0        
    else:
        assert (r + 1) % n != 0
        
    print(n, r)

Which prints:

In other words, it does not halt at an assertion checkpoint! And the conjecture appears true.

Questions.

How can we prove this? Do you already have a proof in mind? Is this a new conjecture?

Original Q&A

There are 2 best solutions below

Bumbble Comm On 04 Feb 2024 - 6:51

The claim is true.

If $n=p_1^{a_1}\cdots p_r^{a_r}$ is a prime factorization of $n$ with $p_1\lt p_2\lt\cdots\lt p_r$ primes and $a_i\geq 1$ for all $i$, then $$\varphi(n) = (p_1-1)p_1^{a_1-1}\cdots (p_r-1)p_r^{a_r-1}.$$ If $n$ is not squarefree, or is even and not $2$, then $\gcd(n,\varphi(n))\gt 1$. In particular, $\varphi(n)!$ cannot be a unit modulo $n$.

If $n$ is squarefree and $r\gt 1$, then $p_1\leq p_2-1\leq \varphi(n)$, so $\gcd(n,\varphi(n)!)\gt 1$. Again, $\varphi(n)!$ cannot be a unit modulo $n$.

The only remaining case is $n$ prime, in which case the congruence is Wilson's Theorem.

**Bumbble Comm** · Accepted Answer

This is a simple argument:

If $n$ is prime, then obviously $$(\varphi(n))! \equiv (n-1)! \equiv -1 \pmod n.$$
Suppose $$(\varphi(n))! \equiv -1 \pmod n$$If any number $k$ in the first $\varphi(n)$ numbers is not coprime to $n$, then this can never be $-1$ modulo $n$. So, we get that $n$ is coprime with all numbers from $1$ to $\varphi(n)$. This means $n$ is not coprime with any number after $\varphi(n)$ (by definition of $\varphi$). Since $(n,n-1) = 1$, $\varphi(n) = n-1$, and the rest follows.

Consider $k \in \{1,2,3 \ldots \varphi(n)\}$. If $\gcd(k,n)>1$, then $(\varphi(n))!$ can't be $-1$ modulo $n$ as that would imply $k$ divides $(\varphi(n))! = xn - 1$ for some $x \in \Bbb Z$, but this is not possible. Since there are exactly $\varphi(n)$ integers in the set $\{1,2,3\ldots , n\}$ coprime to $n$, we know that these integers are exactly the elements of $\{1,2,3 \ldots \varphi(n)\}$. Done.

New conjecture? $(\varphi(n))! = -1 \pmod n \iff n$ is prime (nearly the same as Wilson's)

There are 2 best solutions below

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