Positive integers x,y,z such that $x!+y!=z!$

Question

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?

Answer 1

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?

Answer 2

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 $y=0$.