<s>[INST] Solutions to $x+y+z-2 = (x-y)(y-z)(z-x)$ Show that the equation $$x+y+z-2 = (x-y)(y-z)(z-x)$$ has infinite solutions $(x,y,z)$ with $x, y,z$ distinct integers.
In my attempt to solve the problem only found solutions form $x=y, z=2-2x$.
There are solutions as required by the statement?
 [/INST] Set $x=y+a$ and $z=y-b$. The equation then becomes
$$ 3y+a-b-2 = -ab(a+b) $$
which simplifies to
$$ 3y-2 = -a^2b - ab^2 - a + b $$
If we set $a=2$, then every $b$ that is a multiple of $3$ will lead to a solution for $y$.

In general an integral $y$ is possible exactly when $a+1\equiv b\pmod 3$ or $a\equiv b\equiv 1\pmod 3$, as can be seen by solving
$$ a-b-2 \equiv -ab(a+b) \pmod 3 $$
by brute force, trying each of the 9 possibilities.
This produces every integral soltution.
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