solution of an algebraic equation with integers

Question

Prove that the equation $x^2+y^2+z^2=(x-y)(y-z)(z-x)$ has an infinite number of solutions when $x,y,z$ are integers.

I started from specific cases.

Answer 1

Let $r$ be odd, then $(r+2)^2+r^2+(r+1)^2$ is even, so it is a multiple of $((r+2)-r)(r-(r+1))((r+1)-(r+2))=2$. The quotient is $d=(3r^2+6r+5)/2$. Then $x=d(r+2)$, $y=dr$, $z=d(r+1)$ works.

Of course, there are many more. E.g., let $b=a+1$, $c=a+3$. Then $(a-b)(b-c)(c-a)=6$. If $a$ is even, then $a^2+b^2+c^2$ is divisible by 2, and if $a-1$ is divisible by $3$, then so is $a^2+b^2+c^2$, so if $a=6r-2$, then $a^2+b^2+c^2=108r^2-24r+6=(a-b)(b-c)(c-a)s$ where $s=18r^2-4r+1$. So, $x=(6r-2)s$, $y=(6r-1)s$, $z=(6r+1)s$ is a solution for $r=1,2,3,\dots$.