Thread: 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 .

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

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

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 .

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

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

Adapt Euclid's proof.

Suppose that for some there is no prime such that .

Now consider this is either a prime or composite the first of
these possibilities contradicts our assumption, so it must be composite.

Now let be the primes less than , then clearly .

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

So we conclude that it is not the case that that there are no primes such
that , that is for all there is a prime such that .

RonL

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

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