3
$\begingroup$

In venue of my old question When polynomial f(x^2) can be factored as g(x)·g(-x)? and this recent answer to a different question, I wonder:

How to characterize polynomials $f(x)$ with rational coefficients such that $f(t+t^{-1})=g(t)\cdot g(t^{-1})$, where $g(x)$ is also a polynomial with rational coefficients?

Is there a computationally efficient way to identify if a given polynomial $f(x)$ is such, without factoring $f(t+t^{-1})$ ?

Improve this question
$\endgroup$
4
  • 6
    $\begingroup$ I suspect the two questions may be related via the following substitution (Cayley transform): if $t = \frac{1+x}{1-x}$, then $\frac12 (t+t^{-1}) = \frac{1+x^2}{1-x^2}$. $\endgroup$
    – Federico Poloni
    Commented Feb 12, 2017 at 21:28
  • 1
    $\begingroup$ Another observation derived from similar problems over $\mathbb{C}$ is that $f(\exp(i\theta)+\exp(-i\theta)) = g(\exp(i\theta))g(\exp(-i\theta)) = |g(\exp(i\theta)|^2 \geq 0$, so a necessary condition is that $f(x)\geq 0$ for each $x\in[-2,2]$. $\endgroup$
    – Federico Poloni
    Commented Feb 12, 2017 at 21:34
  • $\begingroup$ @FedericoPoloni: Thanks for the nice observation. It reduces this new question to the old one. Please post your comments as an answer, and I'll accept it (to keep this question for the reference). $\endgroup$
    – Max Alekseyev
    Commented Feb 12, 2017 at 21:43
  • $\begingroup$ I think there is still some work to be done to get a full solution (for instance, to remove denominators and reduce to the polynomial case). It is quite late now in my time zone and I don't have the time to write a complete answer, but if you know how to put everything together please go on and write it yourself -- I will happily leave the imaginary internet points to you. :) $\endgroup$
    – Federico Poloni
    Commented Feb 12, 2017 at 21:50

0

Reset to default

You must log in to answer this question.