# 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 $p \leq n!-1 \leq n!$.

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

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

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 $p \leq n!-1 \leq n!$.

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 $n$ there is no prime $p$ such that $n.

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

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

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

So we conclude that it is not the case that that there are no primes $p$ such
that $n, that is for all $n$ there is a prime $p$ such that $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 $p_i divide n!, they can not divide n!-1.