To find all integers $n > 1$ for which $(n-1)!$ is a zero-divisor in $\mathbb Z_n$.

Question

To find all integers $n > 1$ for which $(n-1)!$ is a zero-divisor in $\mathbb Z_n$. (Gallian Problem)

$\mathbb Z_n$ does not contain any zero divisors when $n$ is a prime number. So we look at the composite numbers.

For $\mathbb Z_4$ we have $(n-1)!$ as $3! =6$ which is equal to $\bar 2$ and it is a zero divisor.

For $\mathbb Z_6$ we have $(n-1)!$ as $5! =120$ which is equal to $\bar 0$ and it is not a zero divisor.

For $\mathbb Z_8$ we have $(n-1)!$ as $7! =5040$ which is equal to $\bar 0$ and it is not a zero divisor.

We Know if $n$ is composite, then $n$ divides $(n-1)!$ when $n \geq 6$.

Thus the only $n$ for which $(n-1)!$ is a zero-divisor in $\mathbb Z_n$ is $n =4$.

Is the solution correct?

Answer 1

0 is a zero-divisor. This means that the statement is true for every composite number by a similar argument to yours. (Reply to comment below: I looked it up on Wikipedia. Appearantly different sources use different definitions. Annoying)

If you change the question to involve non-zero zero divisors, your solution is correct.

However, you may be asked to prove "We Know if n is composite, then n divides (n−1)! when n≥6." It is true, but can you prove it?

Answer 2

If $n$ is a prime then $\mathbb Z_n$ is a field and therefore has no non-zero zero divisors. If $n$ is not a prime we may write $n = ab$ where both $a$ and $b$ are less than $n$. If $a \ne b$, then $(n − 1)!$ includes both $a$ and $b$ among its factors so $(n − 1)! = 0$. If $a = b$ and $a > 2$, then $(n − 1)! = (a^2 − 1)(a^2 − 2)\cdots(a^2 − a)\cdots (a^2 − 2a)\cdots 2 · 1$. Since this product includes $a^2 = n$, it is $0$. The only remaining case is $n = 4$ and in this case $3! = 2$ is a zero divisor.