Find all $(x,y,z)\in\mathbb Z^+$ such that $$\frac{x!+y!}{z!}=3^z$$
I was unable to find a nice way to approach the problem. I tried to use the fact that $\frac{x!+y!}{z!}\equiv 1\pmod 2$ but I got nowhere.I also tried the rearrangement $x!\left(1+\frac{y!}{x!}\right)=z!3^z$ by equating $x!$ and $z!$, and was able to get the solution $(1,2,1)$.
Can someone please provide a rigorous solution providing all solutions or proving no further solutions to this equation.
Here is the solution for $x\geq z$
The equation can be factorised as $$\frac{x!}{z!}(1+\frac{y!}{x!})$$
This implies, $3^{k}=\frac{x!}{z!} (k\leq z)$ or $3^{k}=x(x-1)...(z+1)$ Since $n!$ divides the product of $n$ consecutive integers. $$(x-z)!|3^{k} \rightarrow 0\leq (x-z)\leq 1$$
Case 1: $x=z$
$$3^{x}|1+\frac{y!}{x!}\ \ or\ \ 3^x ∤ (y-x)!$$
So $0 \leq (y-x) \leq 2$ or $y= x,x+1,x+2$ which yields the only solution, $(x,y,z)=(1,2,1)$
Case 2: $x=z+1$
$$3|z\\ \ 3|1+\frac{y!}{(z+1)!}\ \ or\ \ 3 ∤\frac{y!}{(z+1)!}$$
So $1\leq(y-z)\leq 3$ or $y=z+1,z+2,z+3$ from which we get no solutions.
Hence for $x \geq z \ \ \ (1,2,1)$ is the only solution.
Say both $x,y \leq z$
Therefore, max value of the expression is $\frac{2}{z!}$ so $z=2$
It is easy to check that no $x,y$ satisfies this condition.