Find three integer numbers ($x$, $y$, $z$), with $x\ge y\ge z$, so that they comply to the following equations.

Question

I came across the following question on a national math olympiad

Find three integer numbers (x, y, z), with $x\ge y\ge z$, so that they comply to the following equations. $x^2(y-z)+y^2(z-x)+z^2(x-y)=2$

$x+y+z=300$

I solved it in the following manner:

From $x^2(y-z)+y^2(z-x)+z^2(x-y)=2$, when you simplify it you get $(x-y)(y-z)(x-z)=2$ and hence, from the fact that $x\ge y\ge z$, you have that $x-y=1$, $y-z=1$, $x-z=2$. So the solution is $x=101, y=100, z=99$.

My biggest difficulty when solving it was thinking of turning $x^2(y-z)+y^2(z-x)+z^2(x-y)=2$, into $(x-y)(y-z)(x-z)=2$. Can you please explain to me how I could have gone to it more intuitively, so that if I come across a similar problem again, I will immediately know that it can be transformed like that, or to transform something similar into another multiplicative expression?

Answer 1

As for how to have gone about it "more intuitively", here's one option. With $2$ being a prime number, being able to express the left side as a set of factors would greatly reduce the possible number of values. As for what the factors might be, since there is already a $y - z$ factor in the first term, set $d = y - z$ and then check the left side modulo $d$, so $y \equiv z \pmod{d}$, to get

$$x^2(y-z)+y^2(z-x)+z^2(x-y) \equiv z^2(z-x) + z^2(x-z) \equiv 0 \pmod{d} \tag{1}\label{eq1A}$$

This shows $y - z$ is a factor, with symmetry or similar checks showing $z - x$ and $x - y$ are also factors. Then the expansion and collection of terms shows these are the only factors, as you've already determined.