new Fermat's last theorem?

[bob000555]

I have been doing a little fooling around with math(yes I realize how uber nerdy that is) and I may have found a Fermat’s last theorem type problem that is: If x is an integer grater then 1 there is no solution for y^2 = x! or possibly(I doubt it) A^b = x! if y, a, and b are integers. Probably one of the larger factorials proves me wrong but I am just putting it out there.

[Aeternus]

Given for [math]y^2 = x![/math] to hold, the prime factors of all the numbers in the factorial sequence need to even out and until very large numbers (or possibly forever? would have to check), you will always get a new prime [math]q[/math] (an actual prime, not one of the factors of a composite number) after having just got a prime [math]p[/math] before you get [math]p[/math] again as a factor of another number in the factorial sequence.

 

I.E you always (at least until large numbers?) have something of the form -

 

[math]

p ... q ... 2p

[/math]

 

where p and q are primes and therefore won't be able to find such a number where every prime in the sequence of prime factors of the factorial has a twin until this property no longer holds. Perhaps by looking at the formula/approximation for density of primes (perhaps taking some lower bound), one could either show that this always occurs or at least some similar property/extension prevents it, or at least offer a lower bound on the factorial for which it occurs?

[Aeternus]

Yes, so given this http://en.wikipedia.org/wiki/Bertrand%27s_postulate , it seems what I suggested is always the case, and therefore it also seems that there is no solution to y^2=x! for some integers y and x.

 

Given the argument I presented above it seems that this applies to y^n=x! as well (since if you always end up with that odd prime, you can never have 3 of them or 4 of them etc before getting a new odd prime).

 

I imagine this is probably fairly well known and considered quite trivial (either that or I've gotten something wrong but it seems fairly straight forward).