Factoriais!

Originally posted by THUDandBLUNDER
Find all solutions in integers to the following equations:

1) a!b!=a! + b!

2) a!b! = a! + b! + c!

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

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

1) a=2,b=2 (No more solutions because this corresponds to the only solution to xy = x + y in positive integers)

2) If a>b, we must have a! not dividing c!, ie a>c. But then RHS < 3*a!, so b<3. Clearly b can't be 1; if b = 2, we have c! = a! - 2, which has no solutions. So a=b, reducing the equation to a!a! = 2*a! + c!. I can see the solution a=3, c=4, but I can't see a complete solution right now - the best I can do is say that c! + 1 must be a square, but I don't know how many such c there are.

3),4) Don't know - I might have another go later.