finding $2$-cycle in logistic difference equation The logistic difference equation is given by $x_{n+1}=Cx_n(1-x_n)$. I'm supposed to show that when $C>3$, the logistic difference equation has a $2$-cycle.
Letting $f(x)=Cx(1-x)$, if $\{x_0,x_1\}$ is our $2$-cycle we have that $x_{1}=f(x_0)$. Thus, if this is a $2$-cycle, we want to find values of $C$ so that $f^2(x_0)=x_0.$
So what we want is some value of $C$ so that $$f(f(x_0))=Cf(x_0)(1-f(x_0))=C(Cx_0(1-x_0))(1-(Cx_0(1-x_0))) = x_0.$$
My thinking is that, after some manipulation, I should be able to show that this equality can only hold when $C>3$, but actually performing the manipulation which shows that is proving to be a bit of a challenge.
Any thoughts would be greatly appreciated.
Thanks in advance!
 A: Iterating we have:
$$x_{n} = C^{2}x_{n-2}(1-x_{n-2})(1-Cx_{n-2}(1-x_{n-2}))$$
$$x_{n} = C^{2}x_{n-2}(1-x_{n-2})(1-Cx_{n-2} + Cx^{2}_{n-2})$$
Solving for equilibria we have a zero equilibrium and the roots of the polynomial:
$$C^{2} - 1 - C^{2}x - C^{3}x + C^{3}x^{2} + C^{3}x^{2} - C^{3}x^{3} $$
You can factor the known positive equilibrium $\frac{C-1}{C}$ using polynomial long division and then use the quadratic formula to get the periodic solution of period 2.
$$P_{1} = \frac{C + 1 - \sqrt{C^{2} - 2C - 3}}{2C}$$
$$P_{2} = \frac{C + 1 + \sqrt{C^{2} - 2C - 3}}{2C}$$
Which will exist whenever $C^{2} - 2C - 3 > 0$. Assuming $C > 0$ this will occur whenever $C > 3$.
A: To make your life easier, you can do the following:
In order to find the $2$-cycle, you can directly consider the following equation in $x$ [Note the denominator; you don't want the $1$-cycles]
$$\frac{f^2(x) - x}{f(x) - x} = C^2x^2 - C(C+1)x + (C+1) = 0$$
The roots will be real and unequal if the discriminant $$C^2(C+1)^2 - 4C^2(C + 1) > 0 \implies C^2(C+1)(C-3) > 0 \implies C > 3$$ since $C > 0$
Note
The intuition is as follows: if you calculate the second-order fixed points directly, then the first-order fixed points are also calculated. You will get a $4$th degree equation in $x$, ignore the two linear factors corresponding to the two first-order fixed points and consider the quadratic (i.e., the discriminant of the quadratic) in $x$. Here, we avoid these details and directly consider the quadratic. For more details, please see equations $15$-$23$ here
