Find all solutions in positive integers

[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

 

[rasmhop]

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}

 

[xsm113r]

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

 

[xsm113r]

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

Thanks for your help