MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\phi_{n}(x)$ be the $n$-th cyclotomic polynomial. What are the restrictions to $n$ (if any) to have $\phi_{n}(x)$ divides $\phi_{2n}(x)$ (where division is in $\mathbb{Z}[x]$)?Or is it true that $\frac{\phi_{2n}(x)}{\phi_{n}(x)}\in\mathbb{Z}[x]$ for all integers $n$?

share|cite|improve this question
Oh I see...but is it still impossible to have "$\phi_{n}(x)$ divides $\phi_{2n}(x)$" (not necessarily over \mathbb{Z}[x])? – Kikiriku Mar 24 '11 at 13:04
It's just not possible. – Charles Matthews Mar 24 '11 at 15:05
up vote 4 down vote accepted

When is a primitive *n*th root of unity also a primitive 2*n*th root of unity? Please note that the answer is never, and this can also be seen by unique factorisation.

share|cite|improve this answer
Clear...thx.... – Kikiriku Mar 24 '11 at 13:11

Those polynomials are irreducible in $\mathbb Z[X]$ and have different degree... see

share|cite|improve this answer
well, $\phi_{2011}$ and $\phi_{4022}$ have the same degree. – Xandi Tuni Mar 24 '11 at 13:10
Maybe I should think a bit before writing... Of course those polynomials will very often have the same degree (as soon as n is odd)... Kikiriku's answer is much better and does not use the irreducibility of those polynomials... – Aurelien Mar 25 '11 at 13:35

The restrictions are $n$ nonnegative with $n \le 0$. Another characterization is $n=2n.$

I mention that mainly for the humor value. The OEIS comments:

We follow Maple in defining $\Phi_0$ to be $x$; it could equally well be taken to be $1$.

I suppose one could equally well just not define it, a number of sources don't.

$\Phi_{2n}(x)$ is $\Phi_{n}(-x)$ for odd $n$ and $\Phi_n(x^2)$ for even $n$. That would argue that $\Phi_0$ should be $1$ if it is defined.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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