Functional equation which only has constant functions: $f(x)+2f^2(x^2)-1=0,\forall x\in (1,+\infty)$ I need to solve the following functional equation:
$$f(x)+2f^2(x^2)-1=0,\forall x\in (1,+\infty)$$
where $f^2(x^2)$ means the multiplication of the real number $f(x^2)$ with itself.
I sense that any solution of the above functional equation belongs to the class of constant functions on $(1,+\infty)$.
Can anyone suggest something that would help me? I wish happy new year with peace along the world. May only love be in our hearts.
 A: The conjecture is false even if we require the function to be continuous on $(1,+\infty)$. Take $f(x)=-\cos\left(\frac{1}{\ln x}\right)$.
To see that this is a solution, use the double angle formula $\cos x=2\cos^2\frac x2-1$:
$$-f(x)=\cos\left(\frac{1}{\ln x}\right)=2\cos^2\left(\frac{1}{2\ln x}\right)-1=2[f(x^2)]^2-1$$
Hence, $f(x)+2[f(x^2)]^2-1=0$.
Edit: added a proof of its correctness
A: Let me solve this functional equation under some reasonable assumptions. First note that $f(x)=\dfrac{1}{2}, -1$ are the only constant function solutions. Moreover, always $f(x)\le 1$ and equality occures iff $f(x^2)=0$ iff $f(x^4)=\pm\dfrac{\sqrt{2}}{2}.$
Next, note that the given domain can be replace by $(0,\infty)$ with the order reversing bijection $t\mapsto e^{1/2t}$ to transform the given functional equation to $$g(2t)=2(g(t))^2-1$$ for all $t\in (0,\infty),$ where $g(t)=-f(e^{1/2t}).$ Clearly, cosine (on a restricted domain) satisfices the given equation. In fact, for any $a\neq0$ $$g(t)=\cos\left(\dfrac{\pi t}{a}\right)$$ is the only solution under assumptions mentioned in this answer.
