Prove or disprove: Let $f$ be a non-constant polynomial, then $f(x)f(1/x)=1~\Rightarrow~f(x)=\pm x^n,$ for some $n \in \Bbb N$

Question

Prove or disprove: Let $f$ be a non-constant polynomial, then
$$f(x)f(1/x)=1~\Rightarrow~f(x)=\pm x^n,$$ for some $n \in \Bbb N$.

I was trying to prove: If $$f(x)=a_0+a_1x+...+a_nx^n,$$ then $a_0=a_1=...=a_{n-1}=0$ and $a_n=\pm 1$, from the equation $$(a_0+a_1x+...+a_nx^n)(a_0+a_1/x+...+a_n/x^n)=1,$$ I can see this yields $a_0^2+a_1^2+...+a_n^2=1$, then how to reach at $a_0=a_1=...=a_{n-1}=0$ ?

Answer 1

By looking at the coefficient in front of $x^n$, we have $a_na_0 = 0$, but $a_n \neq 0$, so $a_0 = 0$. Then, by looking at $x^{n-1}$, we get $a_na_1 = 0$ due to $a_0$ vanishing, thus $a_1 = 0$, and so on and so forth...
By induction on $\{0,\dots,n-1\}$, you thus get $a_0 = a_1 = \dots = a_{n-1} = 0$, which leaves you with just $a_n^2 = 1$, and thus you have $f = \pm x^n$, which concludes the proof.

Answer 2

Rewrite $f(x) f(1/x)$ as $$\bigg({a_0+a_1x+...+a_nx^n \over x^n}\bigg)(a_0x^n + a_1x^{n-1} + ... + a_n)$$ Now take limits as $x$ goes to infinity...

Answer 3

Take $$(a_0+a_1x+...+a_nx^n)(a_0+a_1/x+...+a_n/x^n)=1$$ and write it as a double sum: $$\sum_{i=0}^n\sum_{j=0}^na_ix^ia_j\frac1{x^j}=\sum_{i=0}^n\sum_{j=0}^na_ia_jx^{i-j}=1$$ If this is true for all $x$ values, all the terms with $i\ne j$ will have to cancel. So let's take the largest power of $x$, $x^n$. This occurs when $i=n$ and $j=0$. So $a_na_0=0$. This can only be true if one of the terms is $0$. Since we assume it's not $a_n$ (otherwise it's not an $n$-th degree polynomial) it means that $a_0=0$. Then we can rewrite the sums above starting with $i$ and $j$ from $1$. Then go to the next power of $x$, which is $n-1$. This occurs when $i=n$ and $j=1$. Therefore $a_na_1=0$. With the same justification, $a_1=0$. And so on, until $a_{n-1}=0$. Then you get to the term with $x^0$, where $a_n^2=1$ (the free term is $1$). This means $a_n=\pm 1$. You don't need to go to lower powers, since there are not any more terms left.

Answer 4

Noting from $f(x)f(1/x)=1$, one has $$ (a_0+a_1x+\cdots+a_nx^n)(a_0x^n+a_1x^{n-1}+cdots+a_{n-1}x+a_n)=x^n$$ So comparing the coefficients of both sides of $x^{2n}$ gives $$ a_0a_n=0 $$ from which one has $$ a_0=0 $$ since $a_n\not=0$. Similarly $$ a_1=a_2=\cdots a_{n-1}=0. $$

Answer 5

$f(x)f(1/x) = 1$ implies that $f(x) = 1/f(1/x)$. So this means that $1/f(1/x)$ is a polynomial in $x$. If $f = a_0 + \ldots + a_nx^n$ then $$ 1/f(1/x) = 1/(a_0 + \ldots + a_nx^{-n})$$ Multiplying the whole thing with $x^n/x^n$ we get $$ x^n / (a_0x^n + \ldots + a_n) $$

Note that as $a_n\neq 0$ nothing cancels out, so this can only be a polynomial in $x$ if $a_0x^n + \ldots + a_n$ is constant in $x$, i.e. if $a_0,\ldots,a_{n-1}=0$.

So we already know that $f(x) = a_nx^n$. Now let us solve $$ (a_nx^n)(a_nx^{-n}) = 1 \Leftrightarrow a_n^2 = 1 \Leftrightarrow a_n = \pm 1$$

This proves your property.

Answer 6

Here is a simple proof for $f\in \mathbb{C}[x]$, i.e. a polynomial over field of complex numbers. Still you can generalize it for any polynomial ring that is a UFD.

Let $n$ be the degree of $f$. Notice that $g(x):=x^nf(1/x)$ is also a polynomial in $\mathbb{C}[x]$, so multiplying both sides of $f(x)f(1/x)=1$ by $x^n$ we get $$f(x)g(x)=x^n.$$

Since $\mathbb{C}[x]$ is a UFD, we must have $f(x)=\alpha x^n,g(x)=\alpha^{-1}$ for some $\alpha \in \mathbb{C}$. Also $g(x)=x^nf(1/x)=x^n\alpha (1/x)^n=\alpha$. Thus $\alpha=\alpha^{-1}$ and so $\alpha^2=1$, which means $\alpha = \pm 1$, as required.