If nN and n2, then the numbers [n!+2,n!+3,n!+4,,n!+n] are all composite. Thus for any n2, we can find n1 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,bN and 1<a,b<c.

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

C={n!+a:aN and 1<an}

For any cC we can write:

c=n!+a=a(n!a+1)=ab, where b=n!a+1

We already know that aN and 1<an<c. All that remains is to show that bN and 1<b<c:

  • b=(n!a+1)N because ana|n!
  • ab=c and 1<a<c, therefore 1<b<c

Therefore any cC 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,,n!+n] the first term is divisible by 2, the second term by 3 and so on. Thus this is a sequence of n1 consecutive composite integers.

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