Solve the functional equation $f(x)f(1/x)=f(x+1/x)+1$, $f(1)=2$, where $f(x)$ is a polynomial. Solve the functional equation
$$f(x)f(1/x)=f(x+1/x)+1,\ f(1)=2,$$
where $f(x)$ is a polynomial.
It is easy to check that $f(x)=x+1$ is a solution. Are there any other solutions? My attempt is described as follows.
Since $f(x)$ is a polynomial, we can write it as $f(x)=\sum_{i=0}^n a_ix^i$, where $a_n\neq 0$. Substituting it into the functional equation, we have
$$
f(x)f(1/x)=\sum_{i=0}^n a_ix^i\sum_{j=0}^n a_jx^{-j}=\sum_{\ell=-n}^n \Bigg(\sum_{i=\max(0,\ell)}^{\min(n,n+\ell)} a_ia_{i-\ell}\Bigg) x^\ell $$
and
\begin{align}
f(x+1/x)+1&=\sum_{i=0}^n a_i(x+1/x)^i+1=\sum_{i=0}^n a_i\sum_{j=0}^i\binom{i}{j} x^{2j-i}+1\\
&=\sum_{\ell=-n}^n \Bigg(\sum_{\max(\ell,0)\leq j\leq \frac{n+\ell}{2}} a_{2j-\ell}\binom{2j-\ell}{j}\Bigg)x^{\ell}+1.
\end{align}
By comparing the coefficients of $x^\ell$ on both sides of the above two equations and noting that $f(1)=\sum_{i=0}^n a_i=2$, we can obtain a very complex system of equations of the coefficients $a_i$, $0\leq i\leq n$. For example, when $n=5$, the system of equations is of the form
\begin{align}
&a_0a_5=a_5,\\
&a_0a_4+a_1a_5=a_4,\\
&a_0a_3+a_1a_4+a_2a_5=a_3+5a_5,\\
&a_0a_2+a_1a_3+a_2a_4+a_3a_5=a_2+4a_4,\\
&a_0a_1+a_1a_2+a_2a_3+a_3a_4+a_4a_5=a_1+3a_3+10a_5,\\
&a_0^2+a_1^2+a_2^2+a_3^2+a_4^2+a_5^2=1+a_0+2a_2+6a_4,\\
&a_0+a_1+a_2+a_3+a_4+a_5=2.
\end{align}
Assuming that $a_5\neq 0$, it is not hard to show that the above system of equations has no solution. But for general case, I have no idea how to solve it or to show it has no solution. I wonder whether there is a clever approach which avoids this tedious computation?
Added. I would like to point out that the condition $f(1)=2$ imposes a strong restriction on the problem. For example, the polynomial $f(x)=1+(1-\sqrt{2})x^2$ satisfies the functional equation but does not satisfy the condition $f(1)=2$. It seems that this condition rules out many solutions.
 A: Here is my solution that was found during an exam :
first let $Deg_p=n>2$ now let :
$$p(x)=\sum_{i=0}^n{a_ix^i}$$
so we have  :
$$p(x)p(\frac{1}{x})=\sum_{i=0}^n{a_ix^i}\sum_{i=0}^n{a_ix^{-i}}=1+\sum_{i=0}^n{a_i(x+\frac{1}{x})^i}$$
first we will look at the constant therm on the both side which is :
$$1+a_0=\sum_{i=0}^n{a_i^2}$$
now we will look at the term $x^n$ at both sides. which gives :
$$a_na_0=a_n$$
since $a_n$ can not be $0$ we must have $a_0=1$ , we rewrite the first condition as :
$$\sum_{i=1}^n{a_i^2}=1$$
now lets calculate the term $x^{n-1}$ on both side we have :
$$a_na_1+a_{n-1}=a_{n-1}$$
which gives $a_1=0$ now we look at the term $n-2$ :
$$a_na_2+a_{n-2}=na_n+a_{n-2}$$
so we have $a_2=n$ but this gives contradiction with the assumption that
$$\sum_{i=1}^n{a_i^2}=1 \ge a_2^2=n^2$$
so we have $n\le 2$ if $n=1$ then we clearly have
$p(1)=2,p(2)=3$ (from $x=1$) so we have $a_1+a_0=2,2a_1+a_0=3$ so $a_1=1,a_0=1$ and $p(x)=x+1$ which clearly works.
if $n=2$ remember we proved $a_1=0$ this also works for $n=2$ so we have
$$p(x)=ax^2+1$$
but since $p(1)=2$ we have $a=1$ bot the polynomial $p(x)=x^2+1$ doesn't work in the statement.
so we are done
A: Concerning the constant term
If the degree $n$ of $f$ is one, there's a unique solution satisfying the functional equation and $f(1)=2$.  For $n\in\{2,3\}$ several functions satisfy the functional equation but not $f(1)=2$.  From now on let $n\geq4$.
We already know that then $f$ has the form
$$f(x)=1+nx^2+\sum_{j=3}^n ajx^j.$$
The constant term of $f(x)f(1/x)$ is the sum of the squares of the coefficients of $f$.  The constant term of $f(x+1/x)$ arises when expanding $(x+1/x)^j$ for even $j$, that is
$$1+2n+\binom{4}{2}a_4+\binom{6}{3}a_6+\binom{8}{4}a_8+\cdots$$
Together with the functional equation we have
$$1^2+n^2+\sum_{j=3}^{n}a_j^{2}=1+2n+\binom{4}{2}a_4+\binom{6}{3}a_6+\binom{8}{4}a_8+\cdots+1.$$
You'll find functions satisfying this equation.  But the function must also satisfy $f(1)=2$, that is
$$1+n+\sum_{j=3}^{n}a_j=2\text{, that is }
\sum_{j=3}^{n}a_j=1-n.$$
