Positive integers x,y,z such that $x!+y!=z!$ Find all positive integers $x,y,z$ such that $$x!+y!=z!$$
My Progress: let $x=y=1$ so that you get $2!$ Which is true. How do I compute this algebraically instead of substituting random values?
 A: In your equation, $x$ and $y$ are interchangeable and one of them has to be smaller or equal to the other one and it doesn't matter which one is larger.
So without loss of generality, let $x\le y$. Clearly for this equation to hold, you need $z > x$ and $z > y$.
Since $x$ is the smallest integer out here, I can divide by it. The idea of dividing by $x!$ is to simplify the equation by the greatest common divisor. Say you want to solve $128x - 512y  = 1024$, you might want to divide by $128$ before anything.
Dividing through by $x!$ gives the following equation :
$$1 +\frac{y!}{x!} = \frac{z!}{x!}$$
This yields that there is only a difference of $1$ between the two integers $\frac{y!}{x!}$ and $\frac{z!}{x!}$. This seems highly improbable for most integers. I will prove that indeed this is only possible in trivial cases (which you already found).  
Recall that if $n>m$, then $\frac{n!}{m!} = n(n-1)(n-2)...(m+2)(m+1)$
First I rearrange and rewrite the factorials as the product of integers (see just above) and deduce the following :
$$1 = z(z-1)(z-2)...(x+2)(x+1) - y(y-1)(y-2)...(x+2)(x+1)$$
Since $z>y$, you can write $\frac{z!}{x!} = \frac{z!}{y!}\frac{y!}{x!}$, a product of integers. So the expression above can be written and factorised as such : $\frac{z!}{x!} - \frac{y!}{x!} = \frac{y!}{x!}\big(\frac{z!}{y!} - 1\big)$.
Expanding the factorial as a long product, I can rewrite it like :
$$1 = \big[y(y-1)...(x+2)(x+1)\big]\big[z(z-1)...(y+1)(y+1) - 1\big]$$
So a product of two positive integers equals one. What can you deduce?
A: Let $x! + y! = z!$ and $x \leq y \ (< z)$
Then $x! + y! \leq 2y!$.
$$z!\leq 2y!$$
$$(y+1)(y+2)...z \leq 2$$
So, $y=1$ or $y=0$.
