# Cyclotomic Polynomials

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 $n$?

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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

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.

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Clear...thx.... –  Kikiriku Mar 24 '11 at 13:11

Those polynomials are irreducible in $\mathbb Z[X]$ and have different degree... see http://en.wikipedia.org/wiki/Cyclotomic_polynomial

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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.

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