Saudi Booklet

Answer: The sequences satisfying the conditions of the problem are:
$$c, -c, c, -c, \dots \\ d, d, d, d, \dots$$
where $c \in \left[-\frac{1}{2}, \frac{1}{2}\right]$ and $d$ is any real number.

We rewrite the first condition as
$$a_{n+2} + a_{n+1} = (a_{n+1} + a_n)(a_{n+1} - a_n + 1)$$
If there exists a positive integer $m$ such that $a_{m+1} + a_m = 0$, then from equation (1) we have $a_{n+1} + a_n = 0$ for all positive integers $n \ge m$. By the fact that the sequence $(a_{i+1} + a_i)$ is periodic, we get $a_{i+1} + a_i = 0$ for every positive integer $i$. Thus the sequence $(a_i)$ is of the form $c, -c, c, -c, \dots$ for some $|c| \le \frac{1}{2}$.

Now suppose that $a_{n+1} + a_n \ne 0$ for every positive integer $n$. Let $T$ be the period of the sequence. From equation (1) we have
$$1 = \prod_{i=1}^{T} \frac{a_{i+2} + a_{i+1}}{a_{i+1} + a_i} = \prod_{i=1}^{T} (a_{i+1} - a_i + 1)$$
Combining with the second condition $|a_{i+1} - a_i| \le 1$, we have $a_{i+1} - a_i + 1 > 0$. Using the AM-GM inequality we get
$$1 = \prod_{i=1}^{T} (a_{i+1} - a_i + 1) \le \left( \frac{\sum_{i=1}^{T} (a_{i+1} - a_i + 1)}{T} \right)^T = 1.$$
So the equality holds, and thus we get
$$a_2 - a_1 = a_3 - a_2 = \dots = a_{T+1} - a_T$$
which means that $(a_i)$ is a constant sequence. So all sequences satisfying the conditions of the problem are those listed above. $\square$