Problem 1767

To prove that there exists a positive integer $p$ such that $u_n = u_{n+p}$ for all $n \in \mathbb{N}$, we will analyze the sequence $(u_n)$ defined by the recurrence relation:
$$u_{n+2} = |u_{n+1}| - u_n$$

We will consider different initial conditions and show that the sequence is periodic with period 9 in all cases.

1. **Case 1: $a = b = 0$**
- If $u_1 = 0$ and $u_2 = 0$, then:
$$u_3 = |u_2| - u_1 = 0 - 0 = 0$$
$$u_4 = |u_3| - u_2 = 0 - 0 = 0$$
$$\vdots$$
- The sequence is $0, 0, 0, \ldots$, which is clearly periodic with period 1.

2. **Case 2: $a = 0$ and $b > 0$**
- Consider $S(0, 1)$:
$$u_1 = 0, \quad u_2 = 1$$
$$u_3 = |u_2| - u_1 = 1 - 0 = 1$$
$$u_4 = |u_3| - u_2 = 1 - 1 = 0$$
$$u_5 = |u_4| - u_3 = 0 - 1 = -1$$
$$u_6 = |u_5| - u_4 = 1 - 0 = 1$$
$$u_7 = |u_6| - u_5 = 1 - (-1) = 2$$
$$u_8 = |u_7| - u_6 = 2 - 1 = 1$$
$$u_9 = |u_8| - u_7 = 1 - 2 = -1$$
$$u_{10} = |u_9| - u_8 = 1 - 1 = 0$$
- The sequence is $0, 1, 1, 0, -1, 1, 2, 1, -1, 0, 1, \ldots$, which is periodic with period 9.

3. **Case 3: $a = 0$ and $b < 0$**
- Consider $S(0, -1)$:
- This sequence is just $S(0, 1)$ shifted by 4 positions:
$$u_1 = 0, \quad u_2 = -1$$
$$u_3 = |u_2| - u_1 = 1 - 0 = 1$$
$$u_4 = |u_3| - u_2 = 1 - (-1) = 2$$
$$u_5 = |u_4| - u_3 = 2 - 1 = 1$$
$$u_6 = |u_5| - u_4 = 1 - 2 = -1$$
$$u_7 = |u_6| - u_5 = 1 - 1 = 0$$
$$u_8 = |u_7| - u_6 = 0 - (-1) = 1$$
$$u_9 = |u_8| - u_7 = 1 - 0 = 1$$
$$u_{10} = |u_9| - u_8 = 1 - 1 = 0$$
- The sequence is $0, -1, 1, 2, 1, -1, 0, 1, 1, 0, -1, \ldots$, which is periodic with period 9.

4. **Case 4: $a \neq 0$ and $b = 0$**
- Using the symmetry property $S(a, b) \iff S(b, a)$, this case is equivalent to Case 2 and Case 3, and thus periodic with period 9.

5. **Case 5: $a \neq 0$ and $b > 0$**
- Consider $S(a, 1)$:
- We need to analyze different ranges of $a$:
5.1) If $a \ge 2$:
$$\overline{a, 1, 1-a, a-2, 2a-3, a-1, 2-a, -1, a-1}, a, 1, \ldots$$
5.2) If $2 \ge a \ge 1$:
$$\overline{a, 1, 1-a, a-2, 1, 3-a, 2-a, -1, a-1}, a, 1, \ldots$$
5.3) If $1 \ge a \ge \frac{1}{2}$:
$$\overline{a, 1, 1-a, -a, 2a-1, 3a-1, a, 1-2a, a-1}, a, 1, \ldots$$
5.4) If $\frac{1}{2} \ge a \ge 0$:
$$\overline{a, 1, 1-a, -a, 2a-1, 1-a, 2-3a, 1-2a, a-1}, a, 1, \ldots$$
5.5) If $0 \ge a \ge -1$:
$$\overline{a, 1, 1-a, -a, -1, 1+a, 2+a, 1, -1-a}, a, 1, \ldots$$
5.6) If $-1 \ge a$:
$$\overline{a, 1, 1-a, -a, -1, 1+a, -a, -1-2a, -1-a}, a, 1, \ldots$$
- In all subcases, the sequence is periodic with period 9.

6. **Case 6: $a \neq 0$ and $b < 0$**
- Since $u_4 = |u_3| - u_2 = |u_3| - b > 0$, the sequence $S(u_3, u_4)$ falls into either Case 2 or Case 5, and thus periodic with period 9.

Hence, in all cases, the sequence $(u_n)$ is periodic with period 9.

$$\boxed{u_{n+9} = u_n \quad \forall n \in \mathbb{N}}$$