Solve $a!b!=a!+b!+c!$ where $a$, $b$ and $c$ are nonnegative integers. My teacher in Math Team gave the following question to us.

Solve $$a!b!=a!+b!+c!$$ where $a$, $b$ and $c$ are nonnegative integers.

I found only one solution by trial and error and it is $(a,b,c)=(3,3,4)$.
Any help is appreciated!
 A: A sketch:
There is no solution with $a\lt2$ (this would imply $c!=-1$) or $a=2$ (this would imply $b!-c!=2$), so by symmetry $a,\,b\ge3$. Without loss of generality $a\ge b$. Since$$a!=\frac{a!}{b!}+1+\frac{c!}{b!}$$is an integer, $c\ge b$.
In the case $a=b$, $b!=2+\frac{c!}{b!}$ is a multiple of $3$, so $c\in\{b,\,b+1,\,b+2\}$. These subcases equate $b!$ to a polynomial in $b$, so only small $b$ can be solutions.
In the case $a>b$, since $b!=1+\frac{b!}{a!}+\frac{c!}{a!}$ is an integer, $a>c\ge b$. The above display-line equation also tells us $\frac{a!}{b!},\,\frac{c!}{b!}$ aren't both multiples of $3$, so $c\in\{b,\,b+1,\,b+2\}$. We can exhaust these using $a!\ge 24,\,b!\ge6$. The case $c=b$ gives $b!=\frac{a!}{a!-2}$, which gives no solutions; $c=b+1$ gives $b!=\frac{a!}{a!-b-2}$, while $c=b+2$ gives $b!=\frac{a!}{a!-b^2-3b-3}$.
A: Let $x,y,z$ be the highest powers of $2$ that divide $a!, b!$ and $c!$, respectively.Then $x+y$ is the highest power of $2$ that divides $a!b!$. 
Also, since you clearly have that $$(a!-1)(b!-1)=c!+1$$ you can obtain that $c>a$ and $c>b$ (apart from some small values of $a,b,c$ which you've probably already ruled out using direct check)
We can now assume WLOG that $a\leq b \leq c$. Then we have:
$$a!+b!+c!=2^xa_1+2^yb_1+2^zc_1=2^x(a_1+2^{y-x}b_1+2^{z-x}c_1)$$
where $a_1,b_1$ and $c_1$ are odd numbers.
Since trivially $x<x+y$ and $2^{x+y}|\text{LHS}$ then we must have that $a_1+2^{y-x}b_1+2^{z-x}c_1$ is even.
If now you assume that $x\neq y$ then $2^{y-x}b_1$ is even, hence we must have $z=x$, i.e. that $c-a=0$ or $c-a=1$ which implies that 
$$(a!)^2\leq a!b!\leq a!+(a+1)!+(a+1)!$$
and that can hold for only small values of $a$ again.
Hence, we must have that $x=y$ hence $b-a=0$ or $b-a=1$.
If $b-a=0$ then we have $$(a!)^2=2a!+c!.$$ As @J.G. showed, this is equivalent to 
$$a!=2+\frac{c!}{a!}$$ hence $c-a=0, 1 \text{ or } 2$ and it reduces to a check for small values.
Finally, if $b-a=1$ then you have $$(a!)^2(a+1)=a!(a+2)+c!$$
which is equivalent to
$$(a+1)!=a+2+\frac{c!}{a!}$$
Now, (unless $c=a$ which again reduces for to small values of a) we have that $a+1| \text{ LHS }$, and $a+1| \frac{c!}{a!}$, hence $a+1|a+2$ which is a contradiction.
Hopefully there aren't any mistakes, though you will have to fill in the gaps, namely do the small values checks.
A: I claim that $a=b$ or $a=c$. For the sake of this problem, assume $b \geq a$ (proving for $b \leq a$ is the exact same). Then
$$b!=1+\frac{b!}{a!}+\frac{c!}{a!}.$$
Due to closure of the integers under addition/subtraction, we get that $c\geq a$.
Case 1: ($b \leq c $)
If $b \leq c$, then
$$b!-1=\frac{b!}{a!}+\frac{c!}{a!}=(\frac{c!}{b!}+1)\frac{b!}{a!}.$$
But $b> a$ implies that $b|(b!-1)$, a contradiction, so $b=a$. Then this equates to solving $(a!)^2=2a!+c!$ or $a!(a!-2)=c!$. For $a\geq 3$, clearly $3 \not|(a!-2)$, so $c=a+1$ or $c=a+2$ (else $c = (a+3)(a+2)(a+1)\dots $ has $3$ as a factor). The equation $a!(a!-2)=(a+1)!$ has one solution at $a=3$ and the equation $a!(a!-2)=(a+2)!$ has none. Then the only solution of this form is $(a,b,c)=(3,3,4)$.
Case 2: ($b>c$)
If $b>c$, then
$$b!-1=\frac{b!}{a!}+\frac{c!}{a!}=(\frac{b!}{c!}+1)\frac{c!}{a!}$$
Because $c<b$, $[c(c-1)(c-2)\dots(c-a+1)] |b!$, whence $[c(c-1)(c-2)\dots(c-a+1)]\not| (b!-1)$ for $c > a$. It follows that $c=a$, which then equates to solving 
$$a!b!=2a!+b!$$
Rearranging both sides gets $a!(b!-2)=b!$, which has no solutions because $b!/(b!-2)$ doesn't take an integer value for $b \geq 3$. Then there are no solutions of the form $(a,b,c) = (a,b,a)$.
Conclusion:
From all of this, we get that $(3,3,4)$ is the only solution to $a!b! = a!+b!+c!$. (Note that I used guessing and checking to prove that $a,b,c > 2$)
