# Diophantine $xy+yz+zx=4(x+y+z)$

How do you solve the Diophantine equation $xy+yz+zx=4(x+y+z)$ for positive integers $x,y,z$? My approach was to consider $d=\gcd(x,y,z)$. I could just about show that the equation has no positive-integer solution for $d=2$, $d=3$ and $d>4$; for $d=4$ there is just one solution $(x,y,z)=(4,4,4)$. However I am stuck with the case $d=1$. How would you tackle this case? Any help appreciated. Thanks.

Write $xy + yz+zx=4x+4y+4z$. Thus they can't be all $>4$. The problem is symmetric. Thus just set $x=0,1,2,3,4$.

• I don't see all of the variables grater than 4... As I said, the problem is symmetric, u get this solution twice, setting x=4 and x=1 and permutating. May 8, 2016 at 13:52

HINT

Here is how you could proceed.

$$z=\frac{-xy+4x+4y}{x+y-4}$$ If $x+y>4$, note that $-xy+4x+4y>0$. This implies that $(x-4)(y-4)<16$

Thus $x \ge 8$ and $y \ge 8$ cannot hold both at once. Assume without loss of generality that $x <8$. There are only a finite number of such $x$.

If $x+y \le 4$, there are only finite such $x,y$.