If and , then the numbers are all composite. Thus for any , we can find 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 can be written as where and .

Proof. Assume is the set of all consecutive numbers following such that:

For any we can write:

We already know that and . All that remains is to show that and :

  • because
  • and , therefore

Therefore any 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 the first term is divisible by , the second term by and so on. Thus this is a sequence of consecutive composite integers.

This one-liner works because all numbers from are divisors of 🤦🏽‍♂️