<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] An easy inspection shows that $f$ is nonconstant. Let $f(x)=x^ng(x)$, with $n\geq0$ and $g(0)\ne0$. Then $f(x)f(1/x)=g(x)g(1/x)$ and $f(x)+f(1/x)=x^ng(x)+\frac{g(1/x)}{x^n}$, hence
$x^ng(x)g(1/x)=x^{2n}g(x)+g(1/x)$. Let $m$ be the degree of $g$, and consider the reciprocal polynomial of $g$, that is $x^mg(1/x)$, who also has degree $m$. We have
$$x^ng(x)x^mg(1/x)=x^{2n+m}g(x)+x^mg(1/x)\,.$$
Note that the LHS has degree $n+2m$, and the summands on the RHS have degrees $2n+2m$ and $m$ respectively. Now $2n+2m=m$ implies $n=0$, and $2n+2m>m$ implies that the degree of the RHS is equal to $2n+2m$, so $2n+2m=n+2m$ and again $n=0$. Thus $f(x)$ has the form $f(x)=a_0+a_1x+\cdots+a_mx^m$, with $m>0$ and $a_0a_m\ne0$, and the equality above becomes $f(x)h(x)=x^mf(x)+h(x)\,,$ where $h(x)=x^mf(1/x)=a_m+a_{m-1}x+\cdots+a_0x^m$. We can rewrite this equality as
$$\bigl(f(x)-1\bigr)h(x)=x^mf(x)\,.$$
Since $0$ is not a root of $h(x)$ then $x^m$ divides $f(x)-1$, and since both $x^m$ and $f(x)-1$ have degree $m$, it follows that $f(x)-1=cx^m$ for some $c\ne0$. Replacing again yields $c(x^m+c)=cx^m+1$, which implies $c=\pm1$. Finally $64=f(4)-1=4^mc$, which in turn forces $c=1$ and $m=3$. We conclude then that $f(x)=x^3+1$.
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