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Just a question I've been thinking about: find/characterize all rational monic polynomials $P(x)$, such that $\frac{P(x^2)}{P(x)}$ is also a polynomial. Of course it is possible that this is not an easy to describe list because of its infinite nature, but I would like to get some assistance in further exploring it.

I already proved that all the roots of $P(x)$ must have magnitude $1$ or $0$ (but those aren't that interesting), or else it would have infinite roots. I have also been able to prove that the roots must have rational (multiples of $\pi$) arguments. I also got a lot of ideas for when $P(x)$ is not constrained to being rational. I'm not sure what I would need to do to also add that constraint. Here are my ideas (they don't constrain $P(x)$ to be rational)

Some examples of polynomials that do work are $P(x)=x^k-1$ and $P(x)=x^{2k}+x^{2k-1}+\ldots+1$ where $k\in\mathbb{N}$. Also some less normal polynomials such as $P(x)=x^4+x^2+1$ and $P(x)=(x-\omega)(x-\omega^2)(x-\omega^4)$ where $\omega=cis\left(\frac{2\pi}{7}\right)$ (I'm not sure if this simplifies to anything nicely). Of course if $P(x)$ and $Q(x)$ both satisfy the condition, then so does $P(x)Q(x)$, so we can combine polynomials together.

At least what I've seen is that if $r$ is a root of $P(x)$, then at the minimum, $r^2$ should also be a root. This is because we need to cancel out the $(x-r)$ term from the denominator with $(x^2-r^2)$ in the numerator. This gets pretty complicated with multiplicity, which I have not been able to completely wrap my head around as $(x^2-r^2)$ can cancel out $2$ roots in the denominator.

Ignoring multiplicity, if we consider a set of the $n$th roots of unity, $\{\omega,\omega^2,\ldots \omega^n\}$, then if we want to remove $\omega^k$, then we have to ensure that there does not exist any $\omega^j$ such that $2j\equiv k\mod n$ (and if we do, we must also remove those roots). This seems like a pretty powerful tool that can obtain the examples I found above and many more.

As for the context of this, I thought about when dealing with the telescoping product $\frac{1}{x^2-x+1}\cdot \frac{1}{x^4-x^2+1}\cdot \frac{1}{x^8-x^4+1}\ldots$. When we use the fact that $x^2-x+1=\frac{x^4+x^2+1}{x^2+x+1}$, the telescoping becomes clear.

Edit: proof of claims about roots having magnitude $1$ and rational multiples of $\pi$ arguments.

AFSOC, that there is a root of $P(x)$ with largest magnitude $r>1$, $re^{i\theta}$. Then as we showed before, $r^2e^{2i\theta}$ must also be a root. However, $r^2>r$ for $r>1$, hence this contradicts with this root having the largest magnitude. Similar reasoning shows that there cannot be a root with magnitude $r<1$ (aside from $r=0$).

My argument for there not being roots with arguments being irrational multiples of $\pi$ is not super rigorous, but if we have $\theta$ being the argument of a root, then $2\theta, 4\theta, 8\theta, \ldots$ must be all be arguments of roots. However, if $\theta$ was an irrational multiple of $\pi$, then its argument will never repeat itself (I'm not sure how to prove this, but it seems reasonable enough).

Alan Abraham
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  • Can you state your proof of "all the roots must have magnitude 1"? EG $P(x) = x$ is a counter example. – Calvin Lin Jan 09 '22 at 16:23
  • Oh, whoops I forgot about that. All the roots must have magnitude $1$ or $0$. I can provide the proof. – Alan Abraham Jan 09 '22 at 16:24
  • Essentially you want $P(x)$ to divide $P(x^2)$. Write them out as product of linear factors over the algebraic closure and see what happens. You can instantly characterise all the roots. – Kolja Jan 09 '22 at 16:30
  • @Kolja I believe that is what I did to figure out that $r$ being a root of $P(x)$ implies $r^2$ being a root of $P(x)$, but I'm not sure what an algebraic closure is. – Alan Abraham Jan 09 '22 at 16:36
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    In the quadratic case, $x^2+bx+c$ gives the solutions $b=-2,-1,0,1$ with $c=b(b+1)/2$ (by long division and cancellation of the remainder). –  Jan 09 '22 at 16:37
  • Let's ignore that. Suppose we work over $\mathbb{C}$. Also, that statement is wrong, if $r$ is a root of $P(x)$ then $\sqrt{r}$ is a root of $P(x^2)$. – Kolja Jan 09 '22 at 16:40
  • @Woody3 I'm pretty sure $x^2-1$ should also be included in that. – Alan Abraham Jan 09 '22 at 16:40
  • @Kolja Sorry, I meant $r^2$ should be a root of $P(x)$. – Alan Abraham Jan 09 '22 at 16:40
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    It seems to me that you're made the progress that you can. The last part could be summarized at looking at equivalance classes of $\mathbb{Z}_n {^\times}$, where $ a \equiv b $ iff $ a = 2^k b \pmod{n}$. Then, these roots will occur with the same multiplicity. EG WIth $n = 7$ , the classes are ${1, 2, 4 }$ and $ { 3, 5, 6 } $. – Calvin Lin Jan 09 '22 at 16:41
  • @CalvinLin While I agree that this works when $n$ is odd, I think this is not true when $n$ is even as we run into issues with this relation not being reflexive e.g. $0=23\pmod 6$ but $\not\exists k\in\mathbb{N}, 3=2^k0\pmod 6$. – Alan Abraham Jan 09 '22 at 16:54
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    Sorry, $b=0$ is a degenerate case, leading to $x^2$ and $x^2-1$. –  Jan 09 '22 at 16:59
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    Does this answer your question? Find the polynomials which satisfy the condition $f(x)\mid f(x^2)$ – more related/duplicate Q&As were pointed out here: https://math.stackexchange.com/q/4050738/42969. (Found with Approach0 – Martin R Jan 09 '22 at 17:00
  • @MartinR Yes, thank you! – Alan Abraham Jan 09 '22 at 17:03
  • @AlanAbraham There are no solutions when $n$ is even, apart from $ 1, -1 $. Sorry, I thought you had dealt with that case already. I was referring to just odd $n$. – Calvin Lin Jan 09 '22 at 19:40
  • @CalvinLin Could you elaborate? $x^4-x^2+1$ works and it has $6$th roots of unity. So does $x^4-1$ and $(x^2-1)(x-i)$. – Alan Abraham Jan 10 '22 at 04:30
  • @AlanAbraham Ah, I misspoke for even $n$. We can have those if their n/2 roots were available. That case needs to be studied. $\quad$ Note that $x^4 - x^2 + 1$ doesn't, neither does $(x-i)$. – Calvin Lin Jan 10 '22 at 16:17

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