If $n\in \mathbb{N}$ and $n\ge 2$, then the numbers $[n!+2,n!+3,n!+4,\dots ,n!+n]$ are all composite. Thus for any $n\ge 2$, we can find $n-1$ consecutive composite numbers. The implication here being that there are arbitrarily large “gaps” between prime numbers.

I came up with an arithmetic proof which utilizes the fact that any composite number $c$ can be written as $c=ab$ where $a,b\in \mathbb{N}$ and $1<a,b<c$.

Proof. Assume $C$ is the set of all consecutive numbers following $n!+1$ such that:

$$C=\{n!+a:a\in \mathbb{N}\text{ and }1<a\le n\}$$

For any $c\in C$ we can write:

$$c=n!+a=a(\frac{n!}{a}+1)=ab\text{, where }b=\frac{n!}{a}+1$$

We already know that $a\in \mathbb{N}$ and $1<a\le n<c$. All that remains is to show that $b\in \mathbb{N}$ and $1<b<c$:

• $b=(\frac{n!}{a}+1)\in \mathbb{N}$ because $a\le n⟹a|n!$
• $ab=c$ and $1<a<c$, therefore $1<b<c$

Therefore any $c\in C$ is composite. $◻$

I realized that I took the bumpy route when I read the wikipedia article on Prime Gaps, which has a much simpler proof.

Proof. In the sequence $[n!+2,n!+3,n!+4,\dots ,n!+n]$ the first term is divisible by $2$, the second term by $3$ and so on. Thus this is a sequence of $n-1$ consecutive composite integers. $◻$

This one-liner works because all numbers from $1\dots n$ are divisors of $n!$ 🤦🏽‍♂️