2005 Indonesia MO Problems/Problem 6

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Problem

Find all triples $(x,y,z)$ of integers which satisfy

$x(y + z) = y^2 + z^2 - 2$

$y(z + x) = z^2 + x^2 - 2$

$z(x + y) = x^2 + y^2 - 2$.

Solution

By using the Distributive Property on all three equations, we get \begin{align*} xy + xz &= y^2 + z^2 - 2 \\ yz + yx &= x^2 + z^2 - 2 \\ zx + zy &= x^2 + y^2 - 2. \end{align*} Add all three equations, rearrange, and factor to get \begin{align*} 2xy + 2yz + 2xz &= 2x^2 + 2y^2 + 2z^2 - 6 \\ 6 &= x^2 - 2xy + y^2 + y^2 - 2yz + z^2 + z^2 - 2xz + x^2 \\ 6 &= (x-y)^2 + (y-z)^2 + (z-x)^2. \end{align*} Notice that since all the variables are integers, the squares must also be integers. The only combination of three integer squares that sum to six is $1, 1, 4$.

Also note that the system is symmetric, so any permutation of a solution will work. Therefore, we can, WLOG, let $x \ge y \ge z$.

Therefore, $(x-y)^2 = 1$, $(y-z)^2 = 1$, and $(x-z)^2 = 4$, so $x-y = 1$, $y-z = 1$, and $x-z = 2$. By rearrangement, $y = x-1$ and $z = x-2$.

Substitute and solve to get \begin{align*} x(x-1) + x(x-2) &= 2x^2 - 6x + 5 - 2 \\ 2x^2 - 3x &= 2x^2 - 6x + 3 \\ 3x &= 3 \\ x &= 1 \end{align*} Therefore, one solution is $(1,0,-1)$. Since any permutation is also a solution, the solutions are $\boxed{(1,0,-1), (1,-1,0), (0,1,-1), (0,-1,1), (-1,0,1), (-1,1,0)}.$