Number of primes from $n!+1$ to $n!+n$

Question

Why aren't there any primes between $n!+1$ and $n!+n$ for all $n>1$? This question was on AHSME 1969 #23, but the question is trivial because it's multiple choice. However, I have no idea how to prove such a statement for infinitely large $n$.

Answer 1

A smaller number that starts $n+1$ consecutive composite numbers is $P(n) =\prod_{p \le n} p $. This is called the primorial of $n$.

To show that $P(n)+i$ is composite for $1 \le i \le n$, just note that each $i$ is divisible by a prime $\le n$, and this prime also divides $P(n)$.

$P(n)$ is much smaller than $n!$, because $\ln(n!) \approx n \ln n - n$ while $\ln(P(n)) \approx n $ (since $\ln P(n)$ is Chebychev's function $\theta(n)$).

Answer 2

This was actually much easier than I had expected. $$n!+i=i(\frac{n!}{i}+1)$$ Also, when $1<i<n$, both factors are integers greater than $1$.

(Credit to vadim123)

Answer 3

For every number $n\in\mathbb{N}$ that you can think of, I can give you a sequence of $n-1$ consecutive numbers, none of which is prime:

• $n!+2$ (divisible by $2$)
• $n!+3$ (divisible by $3$)
• $\dots$
• $n!+n$ (divisible by $n$)

BTW, this proves that there is no finite bound on the gap between two consecutive primes.