A problem about functional equations: $f\left(x^2+\frac14\right)=f(x)$

Question

We want to find all continuous functions $f:\mathbb R\to\mathbb R$ that satisfy the equation $f\left(x^2+\frac14\right)=f(x)$ for all real $x$.
Of course -If I am right- constant functions satisfy the equation mentioned, and as well many other do not such as polynomials, rational functions and etc.; But to find all of them!
I need your hints or solutions with my regards.

Answer 1

From $f(-x)=f((-x)^2+\frac14)=f(x)$ we see that $f$ is even and it suffices to consider $x\ge 0$. For $x_1\in[0,\frac12)$, the sequence defined by $x_{n+1}=x_n^2+\frac14$ is remains within that interval and is strictly increasing, hence converges to the fixpoint $\frac12$. By the property and continuity we conclude $f(x_1)=f(x_2)=f(x_3)=\ldots=f(\frac12)$. For $x_1>\frac12$, the sequence defined by $x_{n+1}=\sqrt{x_n-\frac14}$ remains in $(\frac12,\infty)$, is stictly decreasing, hence converges to the fixpoint $\frac12$. Again we conclude $f(x_1)=f(x_2)=f(x_3)=\ldots=f(\frac12)$.