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 :
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 🤦🏽♂️