SAUDI ARABIAN MATHEMATICAL COMPETITIONS

First, consider 4 sequences $\left(x_n\right),\left(y_n\right),\left(z_n\right),\left(t_n\right)$ with $x_1=|a-b|$, $y_1=|b-c|$, $z_1=|c-d|$, $t_1=|d-a|$, $(a,b,c,d)$ is permutation of $(x,y,z,t)$. And $$\{\begin{array}{l}{x}_{n+1}=\left|a_n-b_n\right|\\ y_{n+1}=\left|b_n-c_n\right|\\ z_{n+1}=\left|c_n-d_n\right|\\ t_{n+1}=\left|d_n-a_n\right|\end{array}$$ with $(a_n,b_n,c_n,d_n)$ is a permutation of $(x_n,y_n,z_n,t_n)$. From this, it is easy to see that $x_n,y_n,z_n,t_n\ge 0$ for all $n$. Let $w_n=\max \left\{x_n,y_n,z_n,t_n\right\},n\ge 1$. We have $$\begin{array}{rl}{w}_n& =\max \left\{x_{n+1},y_{n+1},z_{n+1},t_{n+1}\right\}\\ & =\max \left\{\left|a_n-b_n\right|,\left|b_n-c_n\right|,\left|c_n-d_n\right|,\left|d_n-a_n\right|\right\}\\ & \le \max \left\{a_n,b_n,c_n,d_n\right\}=\max \left\{x_n,y_n,z_n,t_n\right\}=w_n\end{array}$$ The equality occurs when there is at least one number among $a_n,b_n,c_n,d_n$ is equal to $0$. Hence the sequence $(w_n)$ is non-increasing. Suppose that $k$ is a positive integer such that $(x_k,y_k,z_k,t_k)=(x,y,z,t)$, then $w_1\ge w_2\ge \cdots \ge w_k=w_1$ which implies that $w_1=w_2=\cdots =w_k$. So there is at least one number among $(x,y,z,t)$ equal to $0$. Based on the definition of the sequences, we can see that for each $i=1,2,3,\dots ,k$, the tuple $(x_i,y_i,z_i,t_i)$ has some common properties as follows: 1. There is at least one number that is $0$. 2. There are two numbers equal. 3. In the permutation to obtain the next term, two equal numbers must be adjacent. It is easy to check that 3 tuples as follows satisfy the given conditions:

i. If $(x,y,z,t)=(0,0,0,0)$. ii. If $(x,y,z,t)=(a,a,0,0)$ with $a>0$ and its permutation. iii. If $(x,y,z,t)=(2a,a,a,0)$ with $a>0$ and its permutation.

Indeed, when $(x,y,z,t)$ is one of the 3 above tuples, then the form of tuple $\left(x_n,y_n,z_n,t_n\right)$ will not change. Continue, we will prove that if $(x,y,z,t)$ is a tuple that satisfies the condition, then it must be one of the 3 above forms. (*)

Without loss of generality, we may assume that $x\ge y\ge z\ge t=0,x>0$. Since there are two numbers equal among 4 numbers, we consider some cases:

1. If $x=y$ then we have $(x,y,z,t)=(a,a,b,0)$ with $a\ge b\ge 0,a>0$. Based on the 3rd property, $\left(x_1,y_1,z_1,t_1\right)$ can only be $(0,a-b,b,a)$. Apply the 2nd property, we have: - If $a=a-b,b=0$, we have $(x,y,z,t)=(a,a,0,0)$, which satisfies the given condition. - If $a=b,a-b=0$, then we have $(x,y,z,t)=(a,a,a,0)$ with $a>0$. So $\left(x_1,y_1,z_1,t_1\right)$ will be $(0,0,a,a)$. But from this, the form will not change, so we cannot obtain the original form anymore. This case doesn't satisfy. - If $a-b=b$ then $a=2b$, we have $(x,y,z,t)=(2b,2b,b,0)$ and $\left(x_1,y_1,z_1,t_1\right)=(0,b,b,2b)$. But the form of this tuple will not change and similarly, this case doesn't satisfy.

2. If $y=z$, then we have $(x,y,z,t)=(a,b,b,0)$ with $a\ge b>0$. So $(x_1,y_1,z_1,t_1)$ can be $(a-b,0,b,a)$. Apply the 2nd property, we have: - If $a-b=0,a=b$, we have $(x,y,z,t)=(a,a,a,0)$, not satisfy. - If $a-b=a,b=0$, we have the tuple $(x,y,z,t)=(a,0,0,0)$ and $\left(x_1,y_1,z_1,t_1\right)=(a,0,0,a)$. It is easy to see that this case doesn't satisfy. - If $a-b=b$, we have $(x,y,z,t)=(2b,b,b,0)$ which satisfies.

3. If $z=0$ then we have $(x,y,z,t)=(a,b,0,0)$ with $a\ge b\ge 0$. So $(x_1,y_1,z_1,t_1)$ can be $(a-b,0,b,a)$. Apply the 2nd property, we have: - If $a-b=0,a=b$, we have $(x,y,z,t)=(a,a,0,0)$, satisfies. - If $a-b=a,b=0$, we have $(x,y,z,t)=(a,0,0,0)$, not satisfy. - If $a-b=b$, we have the tuple $(x,y,z,t)=(2b,b,0,0)$ and $\left(x_1,y_1,z_1,t_1\right)=(b,b,0,2b)$. It is easy to see that this case doesn't satisfy.

Therefore, there are 3 tuples that satisfy the given condition as above.