Problem 4539, source - "Les mathematiques par les problemes":
Determine all the continuous functions $f:\Bbb R\to\Bbb R$ such that: $$\tag{$1$}\forall x,y\in\Bbb R:f(x+y)f(x-y)=f(x)^2f(y)^2$$
I propose that there are only three continuous solutions $f_{-1,0,1}$, all constant: $$f_k(x)\equiv\begin{cases}-1&k=-1\\0&k=0\\1&k=1\end{cases}$$
Unfortunately this exercise is in the "without solutions" section, so I'd like some comments on the validity or thoroughness of my approach.
My solution:
Let $f$ be any continuous solution to $(1)$. Of course $f_0$ is an easy continuous solution, so suppose $f$ is not identically $0$; it is easy to see then that $f(0)^2=1$ by letting $y=0$ in $(1)$. As a corollary of this, for $f$ nonzero solutions: $$\begin{align}\tag{3}f(2x)f(0)&=f(x)^2\\\tag{4}f(x)f(0)&=f(x/2)^2\\\tag{5}f(-x)f(x)=f(x)^2,\,f(-x)&=f(x)\quad\text{if }f(x)\neq0\end{align}$$Suppose there exists an $\alpha$ with $f(\alpha)=0$. Then $(4)$ gives that $\alpha/2$ is also a root, and inductive application (of $(3)$ also) gives that $2^{-n}\alpha$ is a root for any integer $n$. The supposed continuity of $f$ gives: $$0=\lim_{n\to\infty}f(2^{-n}\alpha)=f(\lim_{n\to\infty}2^{-n}\alpha)=f(0)$$But $f(0)=0$ gives $f\equiv0$ by setting $y=0$ in $(1)$, a contradiction. So, $f$ not identical to $0$ is true if and only if $f$ has no zeroes, so $(5)$ reveals that all non-zero solutions are even. $(4)$ reveals that $f$ is either strictly positive or strictly negative dependent on the sign of $f(0)$ (as $f(0)^2=1,\,f(0)=\pm1$). Then if $f$ is any strictly negative solution, $g:=-f$ is easily seen to be a strictly positive solution. Without loss of generality then we hunt for a positive solution and recover the negative solutions afterward.
Let then $g\gt0$ be a solution of $(1)$. As $g(0)=1$, we have $g(2x)=g(x)^2$ and $g(-x)=g(x)$ for all $x$. Let $x\in\Bbb R$ be arbitrary and fixed. We have: $$g(4x)g(2x)=g(2x)^2g(2x)=(g(x)^2)^3=g(x)^6$$But also via $(1)$: $$\tag{6}g(x)^6=g(4x)g(2x)=g(3x)^2g(x)^2$$We also get: $$g(3x)g(x)=g(2x)^2g(x)^2=g(x)^6$$So that $g(3x)=g(x)^5$ ($g(x)\neq0$ always). By $(6)$: $$g(x)^6=(g(x)^5)^2g(x)^2\implies1=g(x)^6$$For $x$ held arbitrary, and so $g\equiv 1$ as we took $g$ positive.
Thus far $f_0,f_1$ have been identified as the only nonnegative solutions. As mentioned, $f_{-1}:=-f_1$ is then the unique negative solution (technically they have only been shown to be candidate solutions, but that they are continuous and satisfy $(1)$ is obvious). $\blacksquare$
Certainly the solutions I found are correct solutions, but are they all the continuous solutions as I claim?