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.

• As far as I think, the predefined value for $f(1)$ is for us to consider a root of the polynomial rather than defining the polynomial fully, just let $a$ be a root and solve $1+1/x=a$ to get $x=\frac{1}{a-1}$so that $a$ is never $1$ and indeed it's given $1$ is not a zero of the polynomial.
– user732848
Aug 2, 2020 at 17:52
• We have $f(2)=3$. Aug 2, 2020 at 18:00
• Where did you find this problem? Or did you create it? Aug 2, 2020 at 20:02
• Would it help to look at techniques for solving things like $f(x)f(y)=f(x+y)+1$ or $f(x)f(y)=f(x+y)+xy$? Your assumption is weaker (but implied by these when $y=1/x$), but maybe applying a technique for similar equations to your situation might yield ideas. Aug 2, 2020 at 21:44
• Just checked there is no other solution with real coefficients for $\deg f \leq 17$.
– Sil
Aug 12, 2020 at 16:50

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
• I don't get it right: what do you mean by constant term? That one containing no (trivial) power of $x$? I agree, on the left hand side it's the sum of squares of the coefficients of $p$. But on the right hand side some more terms will appear, namely in all $(x+1/x)^i$ where $i$ is even. What am I missing? Jun 17, 2021 at 14:07
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 $$1+n+\sum_{j=3}^{n}a_j=2\text{, that is } \sum_{j=3}^{n}a_j=1-n.$$