Find all continuous functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(x)=f(x^2+C)$ for all $x\in\mathbb{R}$ I posted this question last time but that thread got closed unfortunately because I couldn't write the problem aptly and up to the point of proper understanding.

$4.8.3.$ Let $C>0$ be an arbitrary constant. Find all continuous
functions $f:\mathbb{R}\to\mathbb{R}$ satisfying $f(x)=f(x^2+C),$
for all $x\in\mathbb{R}.$

This question was tried hard by a full group of students but with fail. I request everyone here to try it out.
My ideas: After seeing this problem I was reminded of the famous problem of finding all continuous functions $f:\mathbb{R}\rightarrow\mathbb{R},  f(x^2) = f(x)$, which was easy. I tried the problem in the same and similar manner but eventually failed.
 A: We first consider the case $c \leq 1/4$; we shall show in this case
$f$ must be constant. The relation
$$
f(x) = f(x^2 + c) = f((-x)^2 + c) = f(-x)
$$
proves that $f$ is an even function. Let $r_1 \leq r_2$ be the roots of
$x^2 + c - x$, both of which are real. If $x > r_{2}$, define $x_{0} =
x$ and $x_{n+1} = \sqrt{x_{n} - c}$ for each positive integer $x$. By
induction on $n$, $r_{2} < x_{n+1} < x_{n}$ for all $n$, so the
sequence $\{x_{n}\}$ tends to a limit $L$ which is a root of $x^{2} +
c = x$ not less than $r_{2}$. Of course this means $L = r_{2}$.
Since $f(x) = f(x_{n})$ for all $n$ and $x_{n} \to r_{2}$, we
conclude $f(x) = f(r_{2})$, so $f$ is constant on $x \geq r_{2}$.
If $r_{1} < x < r_{2}$ and $x_{n}$ is defined as before, then by
induction, $x_{n} < x_{n+1} < r_{2}$. Note that the
sequence can be defined because $r_{1} > c$; the latter follows by
noting that the polynomial $x^{2} - x + c$ is positive at $x = c$ and
has its minimum at $1/2 > c$, so both roots are greater than $c$. In
any case, we deduce that $f(x)$ is also constant on $r_{1} \leq x \leq
r_{2}$.
Finally, suppose $x < r_{1}$. Now define $x_{0} = x, x_{n+1} =
x_{n}^{2} + c$. Given that $x_{n} < r_{1}$, we have $x_{n+1} >
x_{n}$. Thus if we had $x_{n} < r_{1}$ for all $n$, by the same argument as
in the first case we deduce $x_{n} \to r_{1}$ and so $f(x) =
f(r_{1})$. Actually, this doesn't happen; eventually we have $x_{n} >
r_{1}$, in which case $f(x) = f(x_{n}) = f(r_{1})$ by what we have
already shown. We conclude that $f$ is a constant function. (Thanks
to Marshall Buck for catching an inaccuracy in a previous version of
this solution.)
Now suppose $c > 1/4$. Then the sequence $x_n$ defined by $x_0 = 0$
and $x_{n+1} = x_n^2 + c$ is strictly increasing and has no limit
point. Thus if we define $f$ on $[x_0, x_1]$ as any continuous
function with equal values on the endpoints, and extend the definition
from $[x_n, x_{n+1}]$ to $[x_{n+1}, x_{n+2}]$ by the relation $f(x) =
f(x^2 + c)$, and extend the definition further to $x < 0$ by the
relation $f(x) = f(-x)$, the resulting function has the desired
property. Moreover, any function with that property clearly has this form.
A: Since $f(-x)=f((-x)^2+C)=f(x^2+C)=f(x)$, one may as well restrict to  $\mathbb R^+=[0,\infty)$.
Let $T:x\mapsto x^2+C$. The topological dynamics of $T$ depends on the value of $C$, which determines the number of fixed points of $T$ in $\mathbb R^+$:    $T$ has zero, one, or two  fixed points, depending on whether $C>1/4$, $C=1/4$, or $C<1/4$.
Here is a sketch of an answer for the case $C>1/4$, when $T$ has no fixed points on $\mathbb R^+$.  Let $t_0=0$ and
define $t_n$ recursively by $t_{n+1}=T(t_n)$.  Note that $\lim_{n\to\infty}t_n=\infty$. The sequence of half-open intervals $I_1=[t_0,t_1),\ldots,I_n=[t_{n-1},t_n),\ldots$ form a partition of $\mathbb R^+$.   Let $g:[t_0,t_1]\to\mathbb R$ be any continuous function such that $g(t_0)=g(t_1)$.  Now let $f(x)=g(x)$ on $I_1$, let $f(x)=g(T(x))$ on $I_2$, and so on, so $f(x)=f(T(x))$ on each $I_n$.
