Find all positive integers a,b,c to the equation a!b!= a!+b!+c! I think I have proved that that a cannot be greater than b (ie a & b must be equal) then using this fact I got a quadratic in terms of a and c, trying out some values I got that (3,3,4) is a solution, but how do I prove this is the only solution or if this is not the only solution then what are the other solutions and how do I find them?
 A: If you prove that a=b, c>a 
$a!(a!-2)=c!$ => $a+1|a!-2$
if $a+1$ is prime, $a+1|a!+1$ =>$a+1|3$ => $a= 2$, but it is not an example
if $a+1 = a_1a_2, 1< a_1 < a_2 < a $, $a+1| a!$ => $a+1|2$ => a=1, but it is not an example  
So, $a+1 = p^2$, p is prime;
if p>2, $a! = C*p*(2p)$, so $a+1| a!$  => as in previous case
So, $a+1 = 2^2$
A: I have a different approach to this. [ETA: I put $x=a, y=b,z=c$ and we assume that $x,y,z$ are POSITIVE integers]
First, assume that $x \le y \le z$.
Next, for each integer $N$ let $\ell(N)$ be the largest power of 3 dividing $N$.

Claim 1: The equation $x=y$ must hold. 

Indeed, suppose otherwise. Then $x,y,z$ satisfy
$$y! = 1 + y!/x! +z!/x!$$
note that if $y>x$ then all but the 1 are divisible by $x+1$, which is impossible.

Claim 2: Either $z < y+3$ or $x=y \le 3$.

Indeed suppose $z \ge x+3$. Then $z! = Bx!$ for some multiple $B$ of 3. Furthermore, $x!+y!+z! = 2x!+Bx!$ [by Claim 1] So $\ell(x!+y!+z!) = \ell(x!)$. However, $\ell(x!y!)= \ell(x!)+\ell(y!)=2\ell(x!)$. Thus by this and Claim 1 $x$ and $y$ must satisfy $x=y<3$ and Claim 2 follows.

Claim 3: If $z$ satisfies $z \le x+2$ then $x \le 5$ and $z\le 7$.

Indeed, by Claim 1 $x=y$, so $x$ and $z$ must satisfy
$$(2 + z!/x!)x! = x! \times x!$$
Plugging in $z \le x+2$ and dividing both sides by $x!$ into the above gives
$$2+(x+1)(x+2) \ge x!$$
This implies $x \le 5$ and thus $z \le 7$.
So now we finish: If $x,y \le 3$ then observe that $z$ satisfies $z \le 6$. So exhaustively check the triplets $(x,y,z); x, y, z \le 7$; $x=y$; $x,y,z$ positive integers.
A: If you have already proved that $a=b$, then
$a!b!=a!+b!+c!$
$(a!)^2=2.a!+c!$
$(a!)^2-2.a!-c!=0$
$(a!-1)^2-c!-1=0$
$c!=(a!-1)^2-1$
So, there is only value for $c$
A: HINT
Let $b \le a, 2 \le a$
$a! = \frac{a!}{b!} + 1 + \frac{c!}{b!}$
So, $a=b$ or $a=b+1$ or $c=b$ or $c = b+1$ (in another case left part is even, but right is odd) 
