Prove that there are no prime numbers in $[n! + 2; n! + n]$

Question

I am struggling trying to prove that there are no prime numbers in $[n! + 2; n! + n]$ gap. Could someone please help?

Answer 1

No prime is in the gap $ [ ] $ $$(n!+2)[ n!+3,n!+4,n!+5,\dots \dots , n!+n-1](n!+n)$$ The number of terms in the sequence is $(n!+n-1)-(n!+3)+1 = n-3$ $$(n!+2)[ n!+3,n!+4,n!+5,\dots \dots , n!+n-1](n!+n)$$ Basically the sequence are all sums of $n!$, since $n!$ is the product of numbers less than it, therefore $ n! +2k$ cannot be prime, $n!+3k$ cannot also be prime,$\dots \dots$, $n!+m\cdot k$ cannot also be prime

What about the case when of $n!+ k$, when $k$ is prime, if $k$ is $\lt n$, means it's multiples are contained in $n!$ then the sum $n!+k$ is not prime

What about $n!+k$, where $k$ is prime and $k \gt n $, here the sum would be a prime number $$(n!+2)[ n!+3,n!+4,n!+5,\dots \dots , n!+n-1](n!+n)$$ In our sequence, the largest value of $k$ is $n-1$, it happens at the last sequence $n!+\{ n-1\} $ but $n-1 \lt n$

Answer 2

All you have to do is realize that if $k< n$ then $k|n!$ so $k|n! + k$ and if $k\ne 1$ then $n! + k$ is not prime.

Answer 3

Odd seeming things happen for $n=1,2$ so I will explicitly restrict my answer to $n>2$

To flesh out the answer given by fleablood: For $2< k \le n$ we have $k>1$, and every number in the interval is of the form $n!+k$ for $n,k$ as defined. Also, since $n!$ has every $k$ as an explicit factor, but not its only factor, every number $\frac{n!}{k}=a_k>1$ is an integer.

Thus for every $k$ as defined, $\frac{n!+k}{k}=(a_k+1) \Rightarrow n!+k=k\cdot (a_k+1)$. Since $n!+k$ has two factors, each greater than $1$, it is not prime, and this is true for every $k$. So no number in the interval is prime.