Polynomial problem

Let $g(x)=f(x)-1$ $\left(g(x)+1\right)\left(g\left(\dfrac1x\right)+1\right)=g(x)+g\left(\dfrac1x\right)+2$ $\Rightarrow g(x)g\left(\dfrac1x\right)=1$ Consider a root $r$ of $g(x)$ . $\lim_{x\rightarrow r} g\left(\dfrac1x\right)$ $=\lim_{x\rightarrow r} \dfrac{1}{g(x)}$ $=\infty$ Since $g\left(\dfrac1x\right)=\dfrac{a_n}{x^n}+\ldots+\dfrac{a_1}{x}+a_0$ is finite for non-zero $x$ , the only root of g(x) must be $0$ .

The only solution is $g(x)=ax^n$ .

Since $g(x)g\left(\dfrac1x\right)=1$ we have $g(x)=x^n$ .

$f(x)=x^n+1$ $f(10)=1001$ $\Rightarrow n=3$ $f(20)=\boxed{8001}$

Another solution:

First note that $f$ cannot be a constant polynomial, since that would result in $f(x)=0,2$ , which violates the given condition. Now, let the degree of $x$ be $n$ . From the given condition, we note that $x^n f(x)=x^nf(1/x)(f(x)-1)=g(x)(f(x)-1)$ where $g(x)$ is another polynomial of degree $n$ . Now, since $f(x)$ has degree at least $1$ , $f(x)-1$ cannot divide $f(x)$ , and we must have $x^n=k(f(x)-1)$ for some scalar $k$ , since the degree of $f$ is also $n$ . This yields that $f(x)=ax^n+1$ , for some $a$ . From the given functional equation, putting this form of $f$ , we get $\left(ax^n+1 \right) \left(\frac{a}{x^n}+1 \right)=ax^n+\frac{a}{x^n}+2\implies a=\pm 1$ If $a=-1$ , then we have $f(10)=-10^n+1=1001\implies 10^n=-1000$ , which is absurd. Thus, $a=1$ , and consequently, $n=3$ . Thus $f(x)=x^3+1$ , which yields the answer as $\boxed{8001}$ .