Prove that if $n\in\mathbb{N}$ and $n \geq 3$ then $n! + 3$ is composite.

Question

Prove that if $n\in\mathbb{N}$ and $n \geq 3$ then $n! + 3$ is composite.

I tried factoring it to show that there are two factors, thus composites but I can't figure out how to get rid of the constant in the equation. Any suggestions for how I could approach this?

Answer 1

Hint

Prove that $3$ divides $n!+3$.

Answer 2

If $3\leq n$ then $n! = 1·2·3·\ldots ·n = 3(1·2·4·\ldots ·n)$ so $n! + 3 = 3(1·2·4·\ldots ·n) + 3 = 3(1+1·2·4·\ldots ·n)$ so $n!+3$ is divisible by $3$.

Answer 3

$$n!=1 \cdot 2 \cdot 3 \cdots n$$

$$n!+3=1 \cdot 2 \cdot 3 \cdots n+3=3(1 \cdot 2 \cdot 4 \cdots n+1)$$

We see that $3 \mid n!+3 \Rightarrow n!+3=3k, k \in \mathbb{Z}$

$n \geq 3 \Rightarrow n! \geq 6 \Rightarrow n!+3 \geq 9, \text{ that implies that } k \geq 3.$

So, $n!+3$ is composite.