Problem 4542, source - "Les mathematiques par les problemes":
Find all the continuous functions $f:\Bbb R\to\Bbb R$ satisfying: $$\tag{1}\forall x,y\in\Bbb R,\,f(x^2-y^2)=f(x)^2-f(y)^2$$
I have made some partial progress with this one, but unlike my previous question, I am not done yet and would like help completing the puzzle. Like the previous question, sadly the book has no solutions for this chapter.
I have determined a few things:
(i) Any solution is necessarily odd
(ii) Any solution must satisfy $f(x)\ge0$ for $x\ge0$ and $f(x)\le0$ for $x\le0$
(iii) Any nonzero solution must satisfy $f(1)=1$
But I have gotten no further.
Proof of (i),(ii):
Let $y=x$ in $(1)$ and one sees $f(0)=0$; letting $y=0$ in $(1)$ and one sees $f(x^2)=f(x)^2\ge0$ so $f(y)\ge0$ for positive $y$. Letting $x=0$ instead gives $f(-y^2)=-f(y)^2\le0$ so that $f(x)\le0$ for all negative $x$, completing (ii). We do also have $\sqrt{f(x^2)}=|f(x)|=|f(-x)|$ and (take $x\ge0$ wlog) $f(x)\ge0,\,f(-x)\le0$ from (ii) so that $f(x)=-f(-x)$ which was to be shown.
I need continuity to show (iii) and the standard result that, for any positive real number $a$, $\lim_{n\to\infty}a^{1/n}=1$.
Let $x\gt0$ be such that $f(x)\neq0$, which is guaranteed by (ii) and a nonzero assumption. From $f(x)=\sqrt{f(x^2)}$ we induct ($f(\sqrt{x})=\sqrt{f(x)}$, etc.) to obtain: $$\tag{2}f(x^{2^{-n}})=f(x)^{2^{-n}}$$Fixing $a:=f(x)\gt0$, we have: $$1=\lim_{N\to\infty}a^{-N}=\lim_{n\to\infty}a^{2^{-n}}=\lim_{n\to\infty}f(x^{2^{-n}})=f(1)$$By continuity.
I am quite at a loss as to how to proceed. I have toyed with ideas like: $$f(x)-f(y)=\frac{1}{f(x)+f(y)}f(x^2-y^2)$$In an attempt to use continuity (bounding $f(x)-f(y)$) but that didn't get anywhere. Continuity also allows me to integrate and try variable substitutions to reduce $(1)$ to some integral where I can use the great result "an integral of a positive continuous function over any interval is $0$ iff. the function itself is $0$" but I couldn't see where to start with that. I personally believe the only solutions are $f\equiv0$ of $f(x)=x$ but I am uncertain how to show this.
Many thanks for any tips/solutions/advice (I have never really done functional equations before :))
N.B. With the classic substitution $x=\frac{1}{2}(a+b),\,y=\frac{1}{2}(a-b)$ for arbitrary $a,b$, and $x^2-y^2=(x+y)(x-y)$, we find from $(1)$ that: $$f(ab)=f\left(\frac{a+b}{2}\right)^2-f\left(\frac{a-b}{2}\right)^2$$But I'm not sure how to apply this either.
From $f(x^2) = f(x)^2$, we get that $f(x^2 - y^2) = f(x^2) - f(y^2)$. Rearrange to get $f(x^2) = f(y^2) + f(x^2-y^2)$. For negative numbers, the oddness of $f$ gives $f(-x^2) = f(-y^2) + f(y^2 - x^2)$. Any pair of real numbers $(a,b)$ can be put in one of those two forms, so $f$ is an additive function. Since $f$ is continuous, it must be of the form $f(x) = cx$. Putting this into the original functional equation gives $c = 0, 1$ as the only solutions.