# a!b! = a! + b! + c!

• Nov 29th 2011, 01:15 PM
jj451
a!b! = a! + b! + c!
This is a BMO 1 question from 2003 that has me completely stumped.

$a!b! = a! + b! + c!$

Any help on basic operations on factorials would be appreciated as I have no idea how they can be manipulated.

Thanks.

EDIT:

Apogies, I forgot to state that all possible values of a b and c are to be found
• Nov 29th 2011, 01:40 PM
Plato
Re: a!b! = a! + b! + c!
Quote:

Originally Posted by jj451
$a!b! = a! + b! + c!$

What is to be done here?
What are the instructions?
• Nov 30th 2011, 12:57 AM
Opalg
Re: a!b! = a! + b! + c!
Quote:

Originally Posted by jj451
This is a BMO 1 question from 2003 that has me completely stumped.

$a!b! = a! + b! + c!$

There is one obvious solution (a,b,c) = (3,3,4). I suspect that there are no others, but I don't see how to prove it.

You could start by rewriting the equation as $(a!-1)(b!-1) = c!+1$, but then what?
• Nov 30th 2011, 01:52 AM
ymar
Re: a!b! = a! + b! + c!
A quick thought (I don't know if it's useful). You can try independent looking at the possible orderings of {a,b,c}. For example, what happens if $a\leq b\leq c$? We can safely assume that $a\leq b$ because of the symmetry in the equation.

With them being ordered, you will know how much you can "divide out" (sorry, don't know the English word) and where.
• Nov 30th 2011, 09:09 AM
jj451
Re: a!b! = a! + b! + c!
If we assume that $a \le b$ say, then we can rewrite $a!b!$ as $(a!)^2 \times \frac{b!}{a!}$ and $a! + b!$ can be rewritten as $2a! + (b! - a!)$ (I know this is obvious but it might help to see something else).

No other ideas at the moment; I'm not too good with factorials.
• Nov 30th 2011, 09:16 AM
alexmahone
Re: a!b! = a! + b! + c!
I found a solution here.
• Nov 30th 2011, 10:20 AM
jj451
Re: a!b! = a! + b! + c!
Quote:

Originally Posted by alexmahone
I found a solution here.

Sehr gut. Marking as solved.