Divisibility of n-1 Factorial by Composite n

Theorem

Let $n \in \Z$ be composite.

Then:

$n \divides \paren {n - 1}! \iff n \ne 4$

where:

$\divides$ denotes divisibility

$n!$ denotes the factorial of $n$.

Proof

Necessary Condition

We have that $3! = 6$ and that $4$ does not divide $6$.

So in order for $n$ to divide $\paren {n - 1}!$ it is necessary that $n \ne 4$.

$\Box$

Sufficient Condition

Let $n \ne 4$ be composite.

Let $n = r s$ where:

$r, s \in \Z_{> 1}$

$r \ne s$

$r, s < n$

This is always possible unless $n = p^2$ for some prime number $p$.

Without loss of generality, let $r < s$.

Then by definition of factorial:

$\paren {n - 1}! = 1 \times 2 \times \ldots \times r \times \ldots \times s \times \ldots \times \paren {n - 2} \times \paren {n - 1}$

and so:

$n = r s \divides \paren {n - 1}!$

Let $n = p^2$.

As $n \ne 4$, we have that $p \ne 2$.

Hence $p > 2$.

Hence:

$2 p < p^2$

and so:

$2 p \divides \paren {n - 1}!$

By definition of factorial:

$\paren {n - 1}! = 1 \times 2 \times \ldots \times p \times \ldots \times 2 p \times \ldots \times \paren {n - 2} \times \paren {n - 1}$

and so:

$n = p^2 \divides \paren {n - 1}!$

Hence the result.

$\blacksquare$

Sources

• 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $4$
• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $4$