TITLE: Find $f(x)$ such that $f(x)+f\left(\frac{1}{x}\right)=f(x)\cdot f\left(\frac{1}{x}\right)$
QUESTION [6 upvotes]: Given a polynomial $f(x)$ of n degree such that
$$f(x)+f\left(\frac{1}{x}\right)=f(x)\cdot f\left(\frac{1}{x}\right)$$
Find the polynomial

I've tried considering $f(x)=\sum \limits_{i=1}^{n} a_{i}x^{i}$ and after a number of steps I finally arrived at the answer.However I am not convinced.Is there a better way to solve this functional equation.
Thanks in advanced!

REPLY [2 votes]: A slight variation of @detnvpp using the fundamental theorem of algebra. Set $g(x)=f(x)-1$ then 
$$g(x)g\left(\frac{1}{x}\right)=(f(x)-1)\left(f\left(\frac{1}{x}\right)-1\right)=f(x)f\left(\frac{1}{x}\right)-f(x)-f\left(\frac{1}{x}\right)+1=1, \forall x\ne 0  $$
if $\alpha \ne 0$ is a zero of the polynomial $g(x)$ then $g(\alpha)g(\frac{1}{\alpha})=0$. That contradicts $g(x)g\left(\frac{1}{x}\right)=1$. Therefore $g(x)$ has no zeros and is constant or $g(x)$ has exactly one zero $\alpha=0$ for. In the first case $g(x)=\pm 1$ and therefore $f(x)=0$ or $f(x)=2$. In the second case $g(x)=x^n$ because of $\left(g(1)\right)^2=1$ and therefore $f(x)=1 \pm x^n$