# Factorial Equation

1. Jul 5, 2012

### FeDeX_LaTeX

1. The problem statement, all variables and given/known data
Find all positive integer solutions for a, b and c to the equation:

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

2. Relevant equations

n! = n(n-1)(n-2)...

3. The attempt at a solution

I'm not having much progress with this. I've tried rewriting it as

(a! - 1)(b! - 1) = c! + 1

and I've found one solution by observation (a = 3, b = 3, c = 4), but I'm not sure what to do. I've ruled out the case a = b = c (since putting that in gives you a! = 3, which has no integer solutions). I've tried comparing the sets of a and b and have tried pairing up numbers by multiplying them to give an element in the set of possible values of c, but this has not worked either, and given that a calculator is not allowed, it would take too long to compute all the possibilities.

A hint (but not a complete solution) would be appreciated here.

EDIT: I've noticed that the remaining solutions all involve a, b and c ≥ 5 (I think). I remembered that, for n > 5 is composite iff (n-1)! = 0 mod n, but I'm not sure how to apply this here?

2. Jul 5, 2012

### micromass

Staff Emeritus
We can assume without loss of generalization that $a\leq b$.

Let's start by a subtle hint. What do you get if you look at that equation modulo 2, modulo 3, etc. ??? Can you draw conclusions about c?

3. Jul 7, 2012

### FeDeX_LaTeX

Well, I've shown that a, b and c cannot take on values 1 or 2 (by exhaustion), so we can say that a, b, c ≥ 3... so the equation is 0 modulo 2 and 0 modulo 3 since both sides are divisible by 3. If we re-arrange;

$$a! = \frac{b! + c!}{b! - 1}$$

Then since a, b, c ≥ 3 the top of the fraction is divisible by 3... but (b! - 1) = 2 mod 3.

4. Jul 7, 2012

### micromass

Staff Emeritus
OK, I'm not really sure how that shows anything...

Let's try in some other way:

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

What do you get mod a! ??

5. Jul 7, 2012

### Saitama

I couldn't do it by algebraic methods. I created a small program to find the values a,b and c. I got only one solution i.e 3,3,4 when i check the values of a,b and c from 1 to 20.
(Please someone rectify if i did something wrong.)

6. Jul 7, 2012

### micromass

Staff Emeritus
That is indeed the only solution. But proving it is an interesting question. If you look at the right divisibility and modulo properties than you should be able to do it.

7. Jul 10, 2012

### FeDeX_LaTeX

Why can you assume without loss of generalisation that $a\leq b$? I used AM-GM on a and b to get $ab \leq \frac{a^2 + b^2}{3}$ but not sure where to go from there...

In mod a!, would it be correct to say that it shows that $b \leq c$? Also not too convinced of this...

Last edited: Jul 10, 2012
8. Jul 10, 2012

### micromass

Staff Emeritus
Otherwise, you just switch a and b around. The situation is completely analogous.

What are x and y?

Not quite. What do you get if you look at the equation modulo a! ?

9. Jul 10, 2012

### FeDeX_LaTeX

Sorry, by x and y I meant a and b, don't know why I used x and y. It's not important now though, I can see why we can assume $a \leq b$ WLOG.

Well, the LHS is b! mod a! ... with the RHS, you get $(2 + c!) \mod a!$ for the case a = b, but for the case b > a, you get the RHS to be $(1 + c! + b(b-1)...(b-(a+1)) \mod a!$, I think.

And the LHS in the case b > a is $b! \mod a!$ OR $a!*b(b-1)...(b - (a+1)) \mod a!$?

Last edited: Jul 10, 2012
10. Jul 10, 2012

### micromass

Staff Emeritus
Isn't b!=0 (mod a!) ??

11. Jul 10, 2012

### FeDeX_LaTeX

How did you get the remainder to be 0...? What happened to the c?

I thought that if a = b then $b \leq c$ as, for the case a = b, you get $b! = (2 + c!) \mod a!$, so $b \leq c$?