# Multiplication of Factorials

1. Dec 28, 2012

### msudidi

Given that a, b, and c are positive integers solve the following equation.

a!b! = a! + b! + c^2

anyone?

2. Dec 28, 2012

### Vorde

I found the answer through brute force: a=2, b=3, c=2.

Not sure if there is a more elegant solution though.

3. Dec 28, 2012

### msudidi

vorde, thanks for trying, im getting the same answer too

but i'm seeking for method to solve it: how to relate multiplication of 2 factorials and their sums? if we can, then c wouldn't be a problem.

there should be a way to solve it

4. Dec 28, 2012

### HallsofIvy

Staff Emeritus
"Brute force" is a method! Please clarify what you are looking for.

5. Dec 29, 2012

### Mandelbroth

Doesn't it work for all positive integers c such that $c = \sqrt{a!b! -a! -b!}$?

Spit-balling here, we have $a!b! = a! + b! + c^2$? Doesn't that imply that $\displaystyle a! = 1 + \frac{a! + c^2}{b!} = 1 + \frac{a(a-1)(a-2)(a-3)...}{b(b-1)(b-2)...} + \frac{c^2}{b!} = 1 + \prod_{k = (b+1)}^a k + \frac{c^2}{b!}$. Don't know where I'm going with that...