TITLE: Find $f(x)$ such that $f(\frac{3}{x})+f(\frac{x}{2})=\frac{9}{x^2}+\frac{x^2}{4}-2$
QUESTION [5 upvotes]: How to find $f(x)$ from below relation ?
$$f\left(\frac{3}{x}\right)+f\left(\frac{x}{2}\right)=\frac{9}{x^2}+\frac{x^2}{4}-2$$ I have two question about this .I find $f(x)=x^2-1$  that work fine ,but
$\\(1)$ how can be sure that there is no other solutions ?
$\\(2)$ How can find $f(x)$ in general ?

REPLY [7 votes]: If $f$ is analytic, we can expand it in a Laurant Series:
$$f(x)=\sum_{n=-\infty}^\infty a_nx^n.$$
Then $f(3/x)+f(x/2)=\sum_{n=-\infty}^\infty 3^na_nx^{-n}+2^{-n}a_nx^n.$
Your functional equation implies that for $n\notin (-2,2,0)$:
$$3^{-n}a_{-n}+2^{-n}a_n=0,$$
You can also work out the coefficients $a_0,a_{\pm 2}$:
$2a_0=-2 \Rightarrow a_0=-1$
$$3^{-2}a_{-2}+2^{-2}a_2=1/4$$
$$3^{2}a_{2}+2^{2}a_{-2}=9,$$
which are the same equation, so $a_2=1-4a_{-2}/9$. So even without the rest of the $a_i$, you can still form an infinite family of solutions:
$$f(x)=\frac{c}{x^2}+(1-4c/9)x^2-1.$$