5
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Tito Piezas III Jan 25 '16 at 14:16
5
$\begingroup$

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: $$ x=3m^2,\quad y=3m^2+1,\quad z=3m^2+1+3m,\quad m\in\mathbb{N}, m\geq 1. $$

$\endgroup$
1
$\begingroup$

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) .

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.