Finding all positive integer solutions to $(x!)(y!) = x!+y!+z!$ The equation is $(x!)(y!) = x!+y!+z! $
where $x,y,z$ are natural numbers.
How to find out them all?
 A: Here's a start. We may assume $x\le y$. The case $x=1$ is trivial. The case $x=2$ gives $y!=z!+2$, which obviously has no solution. So assume $x\ge3$. Then $x!+z!\ge5y!$, so $z!\ge4y!$, so $z\gt y$. Suppose there is a prime number $p$ with $x\lt p\le y$. Then $p$ divides $y!$ and $z!$ but not $x!$ so we get a contradiction from $x!=x!y!-y!-z!$. So we have shown that if $x\le y$ then there can't be a prime $p$ with $x\lt p\le y$. 
A: Without loss of generality, we can assume $x\le y$.  We can quickly rule out any solution with $x=1$ or $2$, since the equations
$$y!=1+y!+z!\quad\text{and}\quad 2y!=2+y!+z!$$
are easily seen to have no solutions in positive integers $y$ and $z$.
Now suppose we had a solution with $x\lt y$.  Then
$$x!={x!\over y!}+1+{z!\over y!}\implies{z!\over y!}=x!-1-{x!\over y!}\not\in\mathbb{N}$$
hence $z\lt y$ as well.  But that implies $x!\lt1+1+1=3$, which implies $x=1$ or $2$, which have already been ruled out.  Hence any solution must have $x=y\ge3$.
Now suppose we have a solution with $x=y\ge3$.  The equation can now be written as
$$x!-2={z!\over x!}$$
which implies $z\ge x$.  We can rule out $z=x$ since $x!$ cannot equal $3$, so we have
$$x!-2=(x+1)(x+2)\cdots(z-1)z$$
But $x\ge3$ implies $x!-2\equiv1$ mod $3$, and this implies the product $(x+1)(x+2)\cdots(z-1)z$ can only be $x+1$ (with $x\equiv0$ mod $3$) -- that is, any product of two or more consecutive numbers is either divisible by $3$ or leaves remainder $2$ mod $3$.  Thus any solution $(x,y,z)$ must be of the form $(x,x,x+1)$, satisfying the equation
$$(x!)^2=x!+x!+(x+1)!$$
which simplifies to
$$x!=x+3$$
which has $x=3$ as its only solution.  So $(x,y,z)=(3,3,4)$ is the only solution to $x!y!=x!+y!+z!$.
A: There won't be very many as the factorials are very sparse.  I suspect you have already listed them all.  We can assume $x \le y$.  If $x!=1$ the RHS is at least $2$ greater than the left.  If $x!=2, y=z$ the RHS is $2$ greater than the left, but there are no factorials with a difference of $2$.  If $x! \ge 6$, divide both sides by $y!$.  The LHS is an integer while the RHS is not unless $x=y$.  Now you are looking for solutions to $(x!)^2=2(x!)+z!$
A: Gerry's argument shows, for $3 \leq x \leq y < z,$ that in fact $x=y.$ Otherwise, if we assume $y \geq x+1,$ there is some prime $p | (x+1),$ which might be $x+1$ itself or might be smaller. Suppose we define $\mbox{ord}_p (n)$ to be the largest exponent $k$ such that $p^k | n.$ We have   $\mbox{ord}_p (y!) \geq 1 + \mbox{ord}_p (x!), $ also  $\mbox{ord}_p (z!) \geq 1 + \mbox{ord}_p (x!).$ Indeed, we have a contradiction, since $ x! = x! y! - y! - z!,$ and everything on the right hand side is divisible by $p^{1 + \mbox{ord}_p (x!)}.$
Alright, so $$x=y.$$ We do, indeed, have the solution $x=3, y=3, z=4.$ 
If $x \geq 4,$ then $\mbox{ord}_2 (x!) \geq 3.$ From $x=y$ we have $ (x!)^2 = 2 x! + z!.$ Now,  $\mbox{ord}_2 (x!)^2 \geq 3 + \mbox{ord}_2 (x!),$ while
 $\mbox{ord}_2 2(x!) = 1 + \mbox{ord}_2 (x!).$ As a result,  $\mbox{ord}_2 (z!) = 1 + \mbox{ord}_2 (x!).$ We cannot have $ z \geq x + 4$ because then there would be two extra even numbers dividing $z!.$ So
$$ z \leq x + 3.$$
However, Stirling's formula shows that, for $x \geq 6, \; z > x + 3$ where we regard $z$ as a real number here. Indeed, for the next few integral $x,$ we get the real solutions $ (4,4,5.833), \; (5,5,7.504), \; (6,6,9.156), \; (7,7,10.814), \; (8,8,12.480), \; (9,9,14.154).$  Stirling's gives me 
$$  z > \; 2 x \; \left(  1 - \frac{\log 2}{\log x} \right)   $$
The only solution in natural numbers is $(x=3, \; y=3, \;z=4)$ 
ADDED: the function I have been calling $\mbox{ord}_p (n)$ is called $v_p(n)$ in Problems from the Book
by Titu Andreescu and Gabriel Dospinescu, available at AOPS or XYZ 
EDIT, December 8::This is sensible notation, as the function $v_p$ is the "$p$-adic valuation," see page 25, Definition 2.1.2, of $p$-adic Numbers by "Fernando Q. Gouvea," that being one of  the pen-names of Mariano Suárez-Alvarez. Mariano wanted to  include a letter "i" somewhere in the last name, have the full set, but was told that would seem inauthentic. P_ADIC_BOOK 
A: Observe that $x, y, z > 2$. Obviously, $y < z$. By symmetry, assume $x \leqslant y$. Then, we have $x \leqslant y < z$. Let $x = y$. Then, $$x! = 2 + z! / x!.$$ Let $k = z - x > 0$, so the last equality becomes $$x! = 2 + (x + 1) (x + 2) \cdots (x + k).$$ Since $x | x!$, we should have $x | k! + 2$. If $k \geqslant x$, then, we get $x | 2$, which is impossible. Hence $k < x$. This is only possible when $k = 1$, in which case we have $x | 3$. Hence, $x = 3$, therefore, $y = 3$ and $z = 4$.
Now suppose $x < y < z$. Let $\alpha$, $\beta$ and $\gamma$ be the largest powers of $2$ that divide $x!$, $y!$ and $z!$, respectively. Then, $$\alpha + \beta = \min (\alpha, \beta, \gamma)$$ by the original equation, since $x, y, z$ are distinct. But this is impossible since $\alpha, \beta, \gamma > 0$.
Hence, the only solution is $(x, y, z) = (3, 3, 4)$.
