TITLE: $f(x)f(1/x)=f(x)+f(1/x)$
QUESTION [16 upvotes]: Find a function $f(x)$ such that:
$$f(x)f(1/x)=f(x)+f(1/x)$$
with $f(4)=65$.
I have tried to let $f(x)$ be a general polynomial:
$$a_0+a_1x+a_2x^2+\ldots a_nx^n$$
which leaves $f(1/x)$ as:
$$a_0+a_1{1\over x}+a_2{1\over x^2}+\ldots + a_n{1\over x^n}$$
On comparing the coefficients of both sides, we see that:
$$2a_0=(a_0)^2+(a_1)^2+(a_2)^2+\ldots+(a_n)^2$$
And
$$a_1=(a_0a_1)+(a_1a_2)+ \ldots +(a_{n-1}a_n)$$
I don't know how to proceed further. I know I need to compare coefficients and come to a conclusion based on their values, but I don't see what to do next.

REPLY [3 votes]: The equation is equivalent to
$$
\frac1{f(x)}+\frac1{f(1/x)}=1
$$
so we can set
$$
f(x)=\frac1{\frac12-\frac{63\,g(\log(x))}{130\,g(\log(4))}}
$$
where $g(x)$ is any odd function.

$g(x)=\tanh(2x)$ gives
$$
f(x)=\frac{11050\left(1+x^4\right)}{10922+128x^4}
$$
Not a polynomial, but at least a rational function.

$g(x)=\tanh\left(\frac32x\right)$ gives
$$
f(x)=1+x^3
$$
which is the answer given by juantheron.