How do you show that the discrete difference equation below has non-trivial solutions of period $2$ for only $r>3$?
$$N_{t+1} = rN_t(1-N_t) $$
Period $2$ solutions satisfy:
$$N_{t+2}=N_t \text{ for all } t$$
and $$ N_{t+1} \ne N_t $$
I found that $N_{t+2} = rN_{t+1}(1-N_{t+1})=r(rN_t(1-N_t)[1-rN_t(1-N_t)]$ but I don't see how this helps.
The solutions for the equation (I use $N \equiv N_t$)
$$ N_{t+2} = r( rN (1-N))(1-rN(1-N)) = N $$ since it's 2-periodic, are:
$$ N = \frac{\pm \sqrt{r^2-2r-3} + r + 1}{2r},\\ N = (r-1)/r $$
The $N = (r-1)/r$ doesnt fit (it's 1-periodic), and since $r^2-2r+3 = (r+1)(r-3)$, it has real solutions for only r > 3.