# 1970 Canadian MO Problems/Problem 1

## Problem

Find all number triples $(x,y,z)$ such that when any of these numbers is added to the product of the other two, the result is $2$.

## Solution

We have: $\begin{matrix} x + yz &=& 2\\ y + xz &=& 2\\ z + yx &=& 2 \end{matrix}$

From the first equation minus the second, we get : $$\begin{matrix}x-y + yz - xz &=& 0 \\(1-z)(x-y) &=& 0\end{matrix}$$

So either $z=1$ or $x=y$. For $z=1$, since the equations were symmetric, we have one solution of $(x,y,z)=(1,1,1)$. From $x=y$, substituting it into the original three derived equations, we have: $$\begin{matrix} x+xz &=& 2\\ z + x^2 &=& 2 \end{matrix}$$ We then get $$z = 2-x^2$$ Substituting this into $x+xz = 2$, $$\begin{matrix} x + x(2-x^2) &=& 2\\ x^3 - 3x + 2 &=& 0\\ (x-1)^2(x+2) = 0 \end{matrix}$$ Thus, either $x = 1$, or $x = -2$. Since the equations were symmetric, we then get the full solution set of: $$(x,y,z) = {(1,1,1), (-2,-2,-2)}$$

## See Also

 1970 Canadian MO (Problems) Preceded byFirst Question 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • Followed byProblem 2
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