4
$\begingroup$

I just found this problem in an old email but I have forgotten how to do it.

Find the number of monic irreducible polynomials $P \in \mathbb{Z}[X]$ such that $P(X) \mid P\left(X^2\right)$ and $\deg(P) = 144$.

$\endgroup$
3
  • $\begingroup$ $X$ and $X-1$ are examples $\endgroup$ – J. W. Tanner Dec 16 '19 at 20:18
  • 1
    $\begingroup$ @J.W.Tanner The problem is about such polynomials with degree 144. $\endgroup$ – Labo Dec 16 '19 at 20:43
  • 1
    $\begingroup$ ... only for odd $n$. For example, $n=292$ does not work even though $\phi(292)=144$. $\endgroup$ – Robert Israel Dec 16 '19 at 20:52
3
$\begingroup$

If $r$ is a root of $P(X)$, it must also be a root of $P(X^2)$, which says $r^2$ is a root of $P(X)$. Thus squaring maps the set of roots of $P(X)$ into itself. This implies that all roots of $P(X)$ must be either $0$ or roots of unity. $P(X)$ is either $X$ or a cyclotomic polynomial. Moreover, you can show that only cyclotomic polynomials $\Phi_j(X)$ with odd $j$ work. Now, what odd $j$ have $\Phi_j(X)$ of degree $144$?

$\endgroup$
4
  • $\begingroup$ Why is it that only odd $j$ work? Using your other observations, I could bruteforce the answer: sagecell.sagemath.org/… $\endgroup$ – Labo Dec 17 '19 at 12:22
  • 1
    $\begingroup$ A root $r$ of $\Phi_j(X)$ is a primitive $j$'th root of unity. If $j$ is even, $(r^2)^{j/2} = 1$, so $r^2$ is a primitive $(j/2)$'th root of unity, not $j$'th. $\endgroup$ – Robert Israel Dec 17 '19 at 12:54
  • $\begingroup$ Oh and when j is odd, 2 is invertible modulo so $r^2$ is still primitive. Nice! $\endgroup$ – Labo Dec 17 '19 at 19:08
  • $\begingroup$ Here is a simpler solution that computes the answer: sagecell.sagemath.org/… $\endgroup$ – Labo Dec 17 '19 at 19:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.