Factorial that are one less than a Square

hey guyz..will u please help me on getting the required of this problem and the solution..i tried but i think i really need your help..

"Are the integers n and x (with n>7) such that n!=x^2-1?

By n! we mean that product of the integers from 1 to n. It is known that 4!+1=25=5^2, 5!+1=121=11^2, and 7!+1=5041=71^2 .

Re: hard algebra question

Hey fireboy25

It should be false but to prove it, you need to have all prime factors to have even power where x^2 = n! + 1. If you can prove that this is not the case then you're done.

Basically find a value of n! where n is prime and check the solution.

Re: hard algebra question

..thank you for your suggestion man ..actually, this question was only given to us by our teacher and does not explained to us anymore..what should i do?..will you please give some informations..

Re: hard algebra question

If n! = x^2 - 1 then n! = (x+1)(x-1). If n! has a prime factor then it means that the result should be n! = prime*(prime-1).

Use this to pick a prime > 7 and show a contradiction if you assume that n! = (x+1)(x-1).

Remember that prime factors can't be factorized further which means that if a prime exists on the LHS, it must also be on the RHS and both sides are simplified as best as they can be (in terms of linear factors).

Re: hard algebra question

This is Brocard's Problem. It is unsolved as of 2012.