Lately I've come up with an interesting math problem to work at. I've been trying to find "factorial triples", or triples of numbers that satisfy $$A!B!=C!$$ with $A,B \ne 1$. By playing around, I've noticed that the relationship is satisfied for all $A$, $B$, and $C$ in the form $$A=K$$ $$B=K!-1$$ $$C=K!$$ And I've also proven the following, under the assumption that some $A,B$ exist satisfying the relationship with $C$:
- If $C-1$ is prime, then $C$ is a perfect factorial.
- If $C-k$ is prime, then some number in the form $$\frac{C!}{(C-n)!}, 1\le n\le k$$ is a perfect factorial.
- C is not prime.
I have also conjectured (but I have not been able to prove) that $C$ cannot be a perfect power of a prime. This is my first question - can anyone give me a hint or two to start this proof?
The thing that bothers me is the triple $$10!=7!6!$$ Because it does not satisfy the form that I had come up with. This one works because $$6!=8\cdot9\cdot10$$ and so I can conclude that if I can find other numbers of the form $$H!=J\cdot(J+1)\cdot...\cdot(J+K)$$ then another possible form is $$A=J-1$$ $$B=H$$ $$C=J+K$$ This is the second question. How can I find such $H,J,K$?
The third and final question is this: how to I prove that all factorial triples are in this form, or, if that is false, how do I find the other forms?
Thanks!