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 by
First Question
1 2 3 4 5 6 7 8 Followed by
Problem 2