<s>[INST] Find $f(x)$ such that $f(x)+f\left(\frac{1}{x}\right)=f(x)\cdot f\left(\frac{1}{x}\right)$ 
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!
 [/INST] $$f(y) = 1+y$$
$$f(x)+f\left(\frac{1}{x}\right) = 1+x+1+\frac{1}{x} = 2 + x + \frac{1}{x}$$
$$f(x) f\left(\frac{1}{x}\right) = (1+x) \left ( 1+\frac{1}{x}\right) = 1 + x + \frac{1}{x} + x \frac{1}{x} = 2 + x + \frac{1}{x}$$
In fact, $f(y) = 1+y^k$ for $k \in \mathbb{N}$ seems to work for the same reason.
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