# 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?

## 3 Answers

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.

• If I may add, more generally, if $a = c+2$ and $b = 3n+c$, then, $$y = \frac{-2 c (1 + c) (2 + c)}{3} - (3 + 8 c + 3 c^2) n - 3(2 + c) n^2$$ However, it can be seen that for any $c = 3m,3m+1, 3m+2$, then $y$ is an integer. – Tito Piezas III Jan 25 '16 at 14:16

Your solution doesn't meet the requirement of the problem because $x=y$.

We can attempt the next obvious parametrization: $x=a$, $y=a+1$. Simplying the original equation using this added info yields: $$2a+z-1=-(a+1-z)(z-a)$$ which can be solved to give: $$z=1-\sqrt{3a}+a\quad\text{or}\quad z=1+\sqrt{3a}+a.$$ Clearly, $a,a+1,a+1+\sqrt{3a}$ are distinct for nonzero $a$. Thus, it remains to select $a$ of the form $a=3m^2$, $m\in\mathbb{N}$, $m\geq 1$, which gives: $$x=3m^2,\quad y=3m^2+1,\quad z=3m^2+1+3m,\quad m\in\mathbb{N}, m\geq 1.$$

The equation $$x+y+z-2=(x-y)(y-z)(z-x)\qquad (*)$$ with three unknowns, has in general infinitely many solutions. There are many ways to get a family of solutions (an incomplete one!).

Look, for example, one particular family of solutions.

$$(*)\Rightarrow \frac{x+y+z-2}{x-y}=(y-z)(z-x)\quad (**)$$ it is allowed any way to achieve that the LHS in $(**)$ is an integer; in particular $x-y=1$. This transforms $(**)$ in $$(y-z)^2+3y-1=0$$ Making now $y-z=t$ one gets $$\begin{cases}y-z=t\\1-3y=t^2\\z=y-t\\x=\frac{1-t^2}{3}+1\end{cases}$$

We chose now $t=3s+1$ and we obtain

$$\begin{cases}x=-3s^2-2s+1\\y=-3s^2-2s\\z=-3s^2-5s-1\end{cases}$$ There are many families of solutions because of the freedom offered by the fact of a single equation with three unknowns. Note that the most of solutions given here are negatives but always in $\mathbb Z$. There are, of course, other parametrizations such that the most of given solutions are positive (such as the example above given by @Kim Jong Un) .