This is a question I made for fun:
Find all continuous $f:(1,\infty)\mapsto\mathbb{R}$ such that $$f(x^2)=f(x)+1$$
I was able to find one solution, namely $$f(x)=\log_2(\log_2x)$$Which I found by first realizing that 1.) $f$ is an even function and 2.) It satisfies: $$f(2^{2^{x}})=x$$ (by fixing $f(2)=0$ and substituting numbers of the form $2^{2^n}$ into the functional equation) then assumed that $f$ is the inverse of $2^{2^{x}}$. I then got an answer. But the question is to find all functions that satisfy the functional equation, and I assumed that $f$ is invertible, which doesn't have to be the case.
I remember seeing the functional equation $f(f(x))=6x-f(x)$ being solved by turning it into $a_2=6x-a_1$ or something like that and then turning the subscripts into powers, so maybe this could also be applied to the functional equation in question?
Best Answer
Claim: all continuous solutions $f:(1,\infty)\to\mathbb R$ take the following form: $$f(x) = g(\log_2 \log_2 x) + \log_2\log_2 x$$ where $g:\mathbb R\to\mathbb R$ is any continuous function of period $1$. Further, the above defines a bijection between the set of continuous functions $(1,\infty)\to \mathbb R$ satisfying the given functional equation, and the set of continuous functions with period $1$. (So you can think of the periodic continuous functions as "parametrizing" solutions to your functional equation.)
Proof. Let $\mathcal C^0(\mathbb T^1)$ denote the set of continuous real-valued functions with period one (i.e. "functions on the 1-torus") and let $F\subset \mathcal C^0(\mathbb R)$ denote the set of continuous functions solving the given functional equation. We will show that the above defines a function $\Phi:\mathcal C^0(\mathbb T^1)\to F$, and moreover that this function is bijective.
First, let $g\in \mathcal C^0(\mathbb T^1)$ be an arbitrary continuous periodic real-valued function. We know that the logarithm base $2$ is defined and continuous on $(0,\infty)$, and that it maps $(1,\infty)$ onto $(0,\infty)$ bijectively, hence $\log_2\log_2 x$ defines a continuous function on $(1,\infty)$. This means that $\Phi(g)$ as defined earlier is a continuous real-valued function $(1,\infty)\to\mathbb R$. Further, we can easily show that it satisfies the required functional equation: \begin{align} f(x^2) &= g(\log_2 \log_2 x^2) + \log_2 \log_2 x^2 \\ &= g(\log_2(2\log_2 x)) + \log_2(2\log_2 x) \\ &= g(\log_2\log_2 x + 1) + \log_2\log_2 x + 1 \\ &= g(\log_2\log_2 x) + \log_2\log_2 x + 1 \\ &= f(x) + 1 \\ \end{align} This shows that $\Phi:\mathcal C^0(\mathbb T)\to F$ is well-defined.
To see that it is injective, suppose that $\Phi(g_1) = \Phi(g_2)$. Then, for any $x\in\mathbb R$, we have that $\Phi(g_1)(2^{2^x}) = \Phi(g_2)(2^{2^x})$, since $2^{2^x} > 1$ for all $x\in\mathbb R$, and therefore $2^{2^x}$ is in fact an element of the domains of these two functions. But substituting the definition of $\Phi$ into this equation yields $$g_1(x) + x = g_2(x) + x$$ and therefore $g_1(x) = g_2(x)$. Since $x$ was arbitrary, we conclude $g_1 = g_2$. Thus $\Phi$ is injective.
Finally we show that $\Phi$ is surjective. Let $f$ be an arbitrary continuous solution to your functional equation, and define $$g(x) = f(2^{2^x}) - x$$ Clearly $g$ is well-defined because $2^{2^x} > 1$ for all real $x$, and it is continuous because $2^{2^x}$ is a continuous function of $x$, and $f$ is also continuous. Further, we can see that $g$ has period $1$, because \begin{align} g(x+1) &= f(2^{2^{x+1}}) - x - 1 \\ &= f((2^{2^{x}})^2) - x - 1 \\ &= f(2^{2^{x}})+1 - x - 1 \\ &= f(2^{2^{x}}) - x \\ &= g(x) \end{align} so we may conclude that $\Phi(g) = f$, hence $\Phi$ surjective. This completes our proof.