jasperE3 wrote: Let $(u_n)$ be a sequence of real numbers which satisfies $$u_{n+2}=|u_{n+1}|-u_n\qquad\text{for all }n\in\mathbb N.$$Prove that there exists a positive integer $p$ such that $u_n=u_{n+p}$ holds for all $n\in\mathbb N$. Let $S(a,b)$ be the sequence $u_1=a$, $u_2=b$ and $u_{n+2}=|u_{n+1}|-u_n$ Note that : a) If $S(a,b)$ is periodic, then $S(ka,kb)$ is periodic with same period whatever if $k>0$ b) All such sequences may be reversed since $u_n=|u_{n+1}|-u_{n+2}$ and so $S(a,b)$ periodic $\iff$ $S(b,a)$ periodic c) note b) implies that periodicity from a given point is equivalent to periodicity from the beginning. So we can limit our study to very few cases : 1) If $a=b=0$ : Sequence is allzero with $1$ being one period. 2) If $a=0$ and $b>0$, enough to study $S(0,1):\overline{0,1,1,0,-1,1,2,1,-1},0,1,...$ indeed periodic with $9$ being one period. 3) If $a=0$ and $b<0$ : Enough to study $S(0,-1)$ which is just $S(0,1)$ shifted four ranks and so periodic with $9$ being one period. 4) If $a\ne 0$ and $b=0$ Using note b) and paragraphs 2) and 3) before, such sequences are also periodic with $9$ being one period. 5) If $a\ne 0$ and $b>0$, enough to study $S(a,1)$ : 5.1) If $a\ge 2$ : $\overline{a,1,1-a,a-2,2a-3,a-1,2-a,-1,a-1},a,1,...$ 5.2) If $2\ge a\ge 1$ : $\overline{a,1,1-a,a-2,1,3-a,2-a,-1,a-1},a,1,...$ 5.3) If $1\ge a\ge\frac 12$ : $\overline{a,1,1-a,-a,2a-1,3a-1,a,1-2a,a-1},a,1,...$ 5.4) If $\frac 12\ge a\ge 0$ : $\overline{a,1,1-a,-a,2a-1,1-a,2-3a,1-2a,a-1},a,1,...$ 5.5) If $0\ge a\ge -1$ : $\overline{a,1,1-a,-a,-1,1+a,2+a,1,-1-a},a,1,...$ 5.6) If $-1\ge a$ : $\overline{a,1,1-a,-a,-1,1+a,-a,-1-2a,-1-a},a,1,...$ And all cases are periodic with $9$ being one period. 6) If $a\ne 0$ and $b<0$ Then, since $u_4=|u_3|-u_2=|u_3|-b>0$, the sequence $S(u_3,u_4)$ is either in case 2), either in case 5) and so periodic $9$ being one period. Hence the claim in all cases $\boxed{u_{n+9}=u_n\quad\forall n\in\mathbb Z_{>0}}$