For $c<1$, let $f_c(x)=x^2+c$, determine the period-2 points?

Question

Ok so I know I need to set $$f_c^2(x)=x$$ so: $$(x_0^2+c)^2+c = x \iff x_0^4+2cx_0^2-x_0+c^2+c=0$$ But how do I then solve this? Ok I have solved for the four set of roots, now you can see there are two roots for $c<\frac{1}{4}$ and four for $c<-\frac{3}{4}$. How do I find the hyperbolicity and therefore the nature of these points?

Answer 1

As user user58697 pointed out $x_0^2 + c - x_0$ and $x_0^4+2cx_0^2-x_0+c^2+c$ are not coprime.

Indeed, $x_0$ is a fix point of $f$ $\iff f(x_0)=x_0\iff x_0^2+c-x_0=0$.

Since fix points of $f$ satisfy $$ f(f(x_0))=x_0 $$ as well, then fix points are also solutions of $$ x_0^4+2cx_0^2-x_0+c^2+c=0. $$ Hence, in this case you may divide $x_0^4+2cx_0^2-x_0+c^2+c$ over $x_0^2 -x_0+ c$ in order to get the polynomial $g(x)$ of (proper) period-2 points of $f$. In this case $g(x)$ is of degree two, so you may solve it.

Answer 2

Notice there is no $x^3$ term.

Try something of the form $(x^2+Ax+B)(x^2-Ax+D)$

$$(x^2+Ax+B)(x^2-Ax+D)=x^4+(D+B-A^2)x^2+(AD-AB)x+BD$$

So we know

$$D+B-A^2=2c$$ $$AD-AB=-1$$ $$BD=c^2+c$$

Which has solution $A=1$, $B=c+1$, and $D=c$.

So $x^4+2cx^2-x+c^2+c=(x^2+x+c+1)(x^2-x+c)$

I assume you can take it from there.