# Thread: Find a prime number

1. ## Find a prime number

Prove that if an integer n > 2, then there exist a prime number p such that n < p < n!

My proof.

Let p be a prime divisor of n!-1, then $\displaystyle p \leq n!-1 \leq n!$.

If p = n, then p|n!, but that is impossible since p|n!-1.

Now, how do I prove that p is not < n?

2. Originally Posted by tttcomrader
Prove that if an integer n > 2, then there exist a prime number p such that n < p < n!

My proof.

Let p be a prime divisor of n!-1, then $\displaystyle p \leq n!-1 \leq n!$.

If p = n, then p|n!, but that is impossible since p|n!-1.

Now, how do I prove that p is not < n?

Suppose that for some $\displaystyle n$ there is no prime $\displaystyle p$ such that $\displaystyle n<p<n!$.

Now consider $\displaystyle N=n!-1$ this is either a prime or composite the first of
these possibilities contradicts our assumption, so it must be composite.

Now let $\displaystyle p_1,\ p_2,\ ..,\ p_r$ be the primes less than $\displaystyle n$, then clearly $\displaystyle p_i \not | N,\ i=1,\ ..,\ r$.

But $\displaystyle N$ is composite, so it has a prime divisor, which therefore must be greater
than $\displaystyle n$ and less than $\displaystyle n!$ , a contradiction.

So we conclude that it is not the case that that there are no primes $\displaystyle p$ such
that $\displaystyle n<p<n!$, that is for all $\displaystyle n$ there is a prime $\displaystyle p$ such that $\displaystyle n<p<n!$.

RonL

3. the part I don't understand is, for the primes less than n, why don't they divide N?

4. For n>2, it's not possible for the divisors of n! to also be the divisors of n!-1 and since all $\displaystyle p_i<n$ divide n!, they can not divide n!-1.