SAUDI ARABIAN MATHEMATICAL COMPETITIONS

First, consider 3 sequences $\left(x_n\right)$, $\left(y_n\right)$, $\left(z_n\right)$, $n\ge 1$ with $$x_1=|x-y|,y_1=|y-z|,z_1=|z-x|\text{ and }\{\begin{array}{l}{x}_{n+1}=\left|x_n-y_n\right|\\ y_{n+1}=\left|y_n-z_n\right|\\ z_{n+1}=\left|z_n-x_n\right|\end{array}$$ From this, it is easy to see that $x_n$, $y_n$, $z_n\ge 0$ for all $n$. Let $w_n=\max \left\{x_n,y_n,z_n\right\}$, $n\ge 1$. We have $$\begin{array}{rl}{w}_{n+1}=\max \left\{x_{n+1},y_{n+1},z_{n+1}\right\}& =\max \left\{\left|x_n-y_n\right|,\left|y_n-z_n\right|,\left|z_n-x_n\right|\right\}\\ & \le \max \left\{x_n,y_n,z_n\right\}=w_n.\end{array}$$ The equality occurs when there is at least one number among $x_n$, $y_n$, $z_n$ equal to $0$. Hence, the sequence $(w_n)$ is non-increasing. Suppose that $k$ is a positive integer such that $(x_k,y_k,z_k)=(x,y,z)$, then $$w_1\ge w_2\ge \dots \ge w_k=w_1\text{ which implies that }{w}_1=w_2=\dots =w_k.$$ On the other hand, for all $i=\overline{1,k}$, there is at least one number among $(x_i,y_i,z_i)$ equal to $0$. Without loss of generality, we may assume that $x\ge y\ge z=0$ and we can see that $x_1=|x-y|$, $y_1=|y-z|=y$, $z_1=|z-x|=x$ so we have some cases:

1. If $|x-y|=0\iff x=y$, we have the tuple $(x,y,z)=(a,a,0)$ with $a\ge 0$. It is easy to check this tuple satisfies the given condition.

2. If $y=0$, to separate from the previous case, we consider $x>0$, we have the tuple $(x,y,z)=(a,0,0)$ with $a>0$. But $x_1=a$, $y_1=0$, $z_1=a$ and from this, we cannot obtain $(a,0,0)$ anymore. Hence, this case does not satisfy the given condition.

3. If $x=0$, we have $x=y=z=0$ which was mentioned above.

Therefore, the tuples $(x,y,z)=(a,a,0)$ and its permutations with $a\ge 0$ satisfy the given condition. $\square$