# Homework Help: Find all solutions in positive integers

1. Dec 29, 2009

### xsm113r

Find all solutions in positive integers a; b; c to the equation
a!b! = a! + b! + c!

I have rearranged and got (a!-1)(b!-1) = c!+1

And the only solutions I can find are a=3 b=3 c=4 but I can't be sure that they are the only ones. How would I go about finding other solutions?

I have tried b=1 to 4 but after that the numbers start getting a bit big

2. Dec 29, 2009

### rasmhop

Re: Factorials

It's clear that we without loss of generality can assume $a \leq b \leq c$. Consider the equation modulo b!. Then you get
$$a! \equiv a! + c! \equiv 0 \pmod {b!}$$
so $b! \, | \, a!$ and therefore you must have $a=b$. Now you can substitute and see $c! = a!(a!-2)$ which you can prove has no solution for a > 3 by noting that if a+1 is prime, then a+1 divides c!, but not a!(a!-2) which is a contradiction; and if a+1=pn is composite with p the smallest possible prime then p and n divide a! so if p and n are not equal a+1=pn|a! and if they are equal, then $a+1=p^2$ so $p,2p \leq a$ for a>3 which proves a+1 | a!. Together this shows $a! \equiv 0 \pmod {a+1}$ whenever a>3 and since $a+1 | a!-2$ we have:
$$2 \equiv a! \equiv 0 \pmod {a+1}$$

3. Dec 30, 2009

### xsm113r

Re: Factorials

Isn't a! + 1 + c! equivalent to 0 (mod b!) though?

4. Dec 30, 2009

### xsm113r

Re: Factorials

Sorry...Just thought about it...I think I get it now.