Integer solutions of $x+y+z=axyz$

Question

let $x,y,z$ be 3 co-prime integers and $a$ a positive even integer. Find all the integer solutions of $$x+y+z=axyz$$ Any hints? What I have done: Suppose there exist $rst\neq 0$ where $$y+z=rx$$ $$z+x=sy$$ $$x+y=tz$$ and $$1+r=ayz$$ $$1+s=azx$$ $$1+t=axy$$ and solve the system below by assuming $\triangle=0$ $$-rx+y+z=0 $$ $$x-sy+z=0 $$ $$x+y-tz=0 $$ $\triangle= r+s+t+2-rst=0 $ Hence, $$a(xy+yz+zx)=1+rst=0$$ I feel like I am going on a circle...

Answer 1

It is easy to deal with the case where $xyz=0$.

If $xyz\not=0$, we have $$a=\left|\frac{x+y+z}{xyz}\right|=\left|\frac{1}{yz}+\frac{1}{zx}+\frac{1}{xy}\right|\le\frac{1}{|yz|}+\frac{1}{|zx|}+\frac{1}{|xy|}\le 3$$ from which we have to have $$a=2$$ So, $$2\le \frac{1}{|yz|}+\frac{1}{|zx|}+\frac{1}{|xy|}\le 3\tag1$$

Supposing here that $$|yz|\ge 2\quad\text{and}\quad |zx|\ge 2\quad\text{and}\quad |xy|\ge 2$$ gives $$\frac{1}{|yz|}+\frac{1}{|zx|}+\frac{1}{|xy|}\le\frac 32$$ which contradicts $(1)$.

Therefore, we have $$|yz|\lt 2\quad\text{or}\quad |zx|\lt 2\quad\text{or}\quad |xy|\lt 2,$$ i.e. $$|yz|=1\quad\text{or}\quad |zx|=1\quad\text{or}\quad |xy|=1$$ I think that you can continue from here.