$f(x+y)f(x-y)=[f(x)f(y)]^2$ implies $f(x)=g\left(x^2\right)$ for some $g$

Question

Small's Functional Equations, Exercise 1.3:

Let $f$ be a real-valued function such that, for all real $x$ and $y$, $$f(x+y)f(x-y)=[f(x)f(y)]^2.$$ Prove that there exists a function $g$ such that $f(x)=g\left(x^2\right)$.

This is from Small's introductory problem book in functional equations which I have been toying with. This question has proved a bit challenging. I was wondering if someone clever could share a hint on which direction this is headed (It's recreational so I'd rather not see a full solution if possible).

I have tried using the angle sum/product trigonometric identities to produce a specific solution but with little luck. I also tried some exponential trial solutions to identify something along that road but as my intuition suggested it appeared not to be the right fit. I am wondering is there is a way to arrive at the conclusion without producing a specific solution. Any thoughts are appreciated!

Answer 1

First of all, note that the existence of such a $g$ is equivalent to the statement that $f(a)=f(-a)$ for all $a$ (since then you can just define $g(x)=f(\sqrt{x})$ for $x\geq 0$).

So now you just want to plug in values of $x$ and $y$ in the functional equation so that you will get statements relating $f(a)$ and $f(-a)$. So for instance, you probably want to choose $x$ and $y$ such that one or more of $x+y$, $x-y$, $x$, and $y$ are equal to $a$ or $-a$.

A full solution is hidden below.

First, setting $y=0$ gives $f(x)^2=f(x)^2f(0)^2$. This means $f(0)^2=1$ unless $f(x)=0$ for all $x$. The latter case is trivial, so let us assume $f$ is not identically $0$ and so $f(0)^2=1$. Now set $x=0$ and $y=a$ to get $$f(a)f(-a)=f(0)^2f(a)^2=f(a)^2.$$ Unless $f(a)=0$, this implies $f(a)=f(-a)$, as desired. If $f(a)=0$, we need to do a little more work: setting $x=0$ and $y=-a$ gives $$f(-a)f(a)=f(0)^2f(-a)^2=f(-a)^2.$$ Since $f(a)=0$, this implies $f(-a)=0$ as well, and so we're done.