If $\{x_n\}$ is a real sequence with period $T$ obeying $x_{n+2} = 1+x_{n+1} x_n,$ what are the possible values of $T$?
$T=3$ is possible: $2, -1, -1, 2, \dots$. In fact, some analysis shows this to be the only example.
I have ruled out $T=2, 4$ after some work, and $T=1$ does not work since the roots of $x^2+1=x$ are non-real.
Does anyone have any ideas about non-brute force methods for ruling out or finding higher values of $T$?
Edit: I have determined $0$ cannot be part of a periodic sequence, and a periodic sequence must have sign pattern $+--+--\dots,$ ruling out all periods that are not multiples of $3.$ First note $++$ makes the sequence unbounded, so $+$ is followed by $-.$ This also means $-+$ is followed by $-.$ A lengthy analysis of magnitudes reveals that $+-+$ is impossible, so $+-$ is followed by $-.$ Finally, $--$ is clearly followed by $+.$ If $0, a, \dots, b, 0, a, \dots$ is periodic, then $a = b \cdot 0 + 1 = 1,$ so the sequence begins $0,1,$ which does not work.
Multiplying by $x_{n-1}$ we get $$x_{n-1}x_{n+2}=x_{n-1}+x_{n-1}x_{n}x_{n+1}.$$ Multiplying by $x_{n+2}$ we get $$x_{n+2}^2=x_{n+2}+x_{n}x_{n+1}x_{n+2}.$$ If we sum equalities of both types over all $n=1,2,\dots T$, we get that $$\sum x_{n}x_{n+3} = \sum x_{n}^2, $$ or $$\sum (x_{n}-x_{n+3})^2=0.$$ Hence $T|3$, but we know that $T\not = 1$, so $T=3$.