<s>[INST] If $f(x) $ be a polynomial function satisfying $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$ and $f(4) =65$ then find $f(6)$ Problem : 
If $f(x) $ be a polynomial function satisfying $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$ and $f(4) =65$ then find $f(6)$ 
Solution : 
$f(x) f(\frac{1}{x})-f(x) =f(\frac{1}{x})$
$\Rightarrow f(x) =\frac{f(1/x)}{f(1/x)-1}$.....(i)
Also $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$
$\Rightarrow f(\frac{1}{x})=\frac{f(x)}{f(x)-1}$ ......(ii)
On multiplying (i) and (ii) , we get 
$f(x) .f(\frac{1}{x})=\frac{f(1/x).f(x)}{(f(1/x)-1) ((f)(x)-1)}$
$\Rightarrow (f(\frac{1}{x}) -1)(f(x)-1)=1$ 
Please suggest how to proceed here since $f(x)-1 ; \& f(\frac{1}{x}-1)$ are reciprocal to each other 
Thanks
 [/INST] Let $f$ be an integer polynomial with $f(x)f(1/x) = f(x) + f(1/x)$. There are four solutions, $f = 0$, $f = 2$, and $f = 1 \pm x^n$. 
Clearly $f = 0$ is a solution.
As in the question, we have:
$$ (f(x) - 1)(f(1/x) - 1) = 1 $$
Let $n$ be the degree of $f \neq 0$. Multiplying both sides of the above by $x^n$ gives:
$$ (f(x) - 1)(x^n f(1/x) - x^n) = x^n $$
Now both terms are now polynomials in $\mathbb Z [x]$ with degree $n$. Therefore, by unique prime factorization one of the factors on the left must be $x^n$ (up to units), and the other factor is some unit. This gives two possibilities, either:
$$ f(x) - 1 = \pm x^n $$
$$ f(x) = 1 \pm x^n $$
The other possibility:
$$ x^n f(1/x) - x^n = \pm x^n$$
$$ f(1/x) = 0, 2 \implies f(x) = 2 \text{ since $f \neq 0$}$$

Applying this to your question, we get two possibilities for $f(x)$. Either:
$$ 4^n + 1 = 65 $$
$$ 4^n - 1 = 65 $$
Since $n$ has to be a natural number, the only solution is $n=3$, giving the polynomial $f(x) = x^3 + 1 $.
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