Showing a number $n$ is prime if $(n-2)! \equiv 1 \pmod n$

Question

I need to show that if $(n-2)! \equiv 1 \pmod n$ then $n$ is prime.
I think that if $n$ is composite, then we'll have every factor of $n$ in $(n-2)!$, and it would yield that $(n-2)! \equiv 0 \pmod n$.
However, I didn't use the fact that it specifically congruent to $1 \bmod n$, so I think I'm getting something fundamental wrong. Is my solution correct? Why do we demand congruence to $1 \bmod n$?

Answer 1

Your solution is correct aside from the small detail of 4!, which has already been pointed out in the comments. 4 is somewhat special, being the first composite number, and the square of the "oddest" prime. So I don't think this little mistake is at all "fundamental."

Still, I think it's better to use the fact of $1 \bmod n$, since it guarantees that $\gcd((n! - 2), n) = 1$. The least prime factor of $n$, if $n$ is not itself prime, is less than or equal to $\sqrt n$, and $\sqrt n < (n - 2)$ for all $n > 4$. Therefore, if $n$ is composite, then $(n - 2)!$ is divisible by the least prime factor of $n$, making $(n - 2)! \equiv 1 \bmod n$ impossible.

Yet another way you can go about it is this: If $n$ is even and composite, then $n - 2$ is also even and therefore $(n - 2)! \equiv 2k \bmod n$, where $k$ is some nonnegative integer we don't care too much about. But 1 is not even, proving $n$ is not an even composite number. If $n$ is odd and composite, it is divisible by some odd prime $p \leq \sqrt n$. And since $p \leq \sqrt n < (n - 2)$, it follows that $p \mid (n - 2)!$ and $(n - 2)! \equiv pk \bmod n$. And $pk \neq 1$, proving $n$ is not an odd composite number.

And yet another way is to use Wilson's theorem, but then I would be merely restating one or two of the other answers.

Answer 2

If $n=p^2$ where $p>2$ is a prime number, $(n-2)!$ contains $p$ and $2p$ inside it, so it is $0$ modulo $p^2$. If $n=2^2$, then $(n-2)!\equiv 2$. In all other cases it is easy to see that composite $n=ab$ where $a\ne b$, and then $(n-2)!$ contains both $a$ and $b$ and is $0$ modulo $n$. If $p$ is prime, then $(-1)(n-2)!\equiv(n-1)!\equiv -1$ by Wilson's theorem and so $(n-2)!\equiv 1$.

So generally you use the fact that $(n-2)!$ is not $0$ or $2$, it is not necessary to know that it is exactly $1$. Just cause it cannot be anything except $0,1,2$.

Answer 3

As $\mathbb{Z}_n$ is a commutative unitary ring, all we need to prove is that every element has inverse, so it'd be a field and then $n$'d be prime. Is easy to see from the hypothesis than for any $k\leq n-2$ we have the inverse. Then we just note that $-(n-2)!\mod n$ is the inverse of $(n-1)$, 'cause: $$-(n-1)(n-2)!=-(n-1)\dot{}1=1\mod n$$

Answer 4

Let $n = a*b; n > 4$ and neither $b$ nor $a$ is $1$.

First off: both $b$ and $a$ are less than $n-2$. This should be clear when you consider if $b \ge n-2$ then $a = \frac nb \le \frac n{n-2} < \frac n{n- \frac 12 n} = 2$ so $a < 2$ which contradicts our hypothesis. (Note: we need $n > 4$ to know that $n- 2 > n- (\frac 12 n)$)

If $a \ne b$ then $(n-2)! = \prod_{k \le n-2;k \in \mathbb N} k$ and $\frac {(n-2)!}{ab} = \prod_{k \le n-2;k \ne a; k\ne b;e \in \mathbb N}k \in \mathbb Z$ so $(n-2)! \equiv 0 \mod ab = n$.

If $a = b$ and $n = a*b$ is the only possible factorization then $n = a^2$ and $a$ is prime.

But as $n > 4$ we have $a > 2$ so we have $a < 2a < a^2 - 2 < a^2$ so $\frac{(a^2 - 2)!}{a^2} = 2*\prod_{k\le a^2 -2; k \ne a; k \ne 2a k \in \mathbb N}k$.

So $(n-2)! \equiv 0 \mod n$ if $n$ is composite.

So if $(n-2)! \not \equiv 0 \mod n$ for $n > 4$ then $n$ must be prime.

Of course, we haven't proven that $(p-2)! \ne 0 \mod p$ (although that's obvious) or that there are any $(p-2)! \equiv 1 \mod p$ (not as obvious).