2
$\begingroup$

Let $\Phi_n(t)$ be the $n$th cyclotomic polynomial. It dates back to Dirichlet that $\Phi_m(t)$ and $\Phi_n(t)$ are coprime over ${\mathbb Q}$ for $m\ne n$. Therefore there exist integer polynomials $a(t),b(t)\in{\mathbb Z}[t]$ and a positive integer $d$ such that $$a(t) \Phi_m(t)+b(t)\Phi_n(t)=d.$$

Question: If $\frac nm$ is not a prime power, then can $d$ be chosen to be 1?

Any pointers to the literature, or a nice proof would be appreciated. (By a "prime power" I mean a positive integer power of a prime.)

$\endgroup$
3
$\begingroup$

Yes: If $\dfrac{n}{m}$ is not a prime power, then we can find $a\left(t\right)$ and $b\left(t\right)$ in $\mathbb{Z}\left[t\right]$ satisfying $a\left(t\right) \Phi_m\left(t\right) + b\left(t\right) \Phi_n\left(t\right) = 1$. This follows from the fact that the resultant of the polynomials $\Phi_m\left(t\right)$ and $\Phi_n\left(t\right)$ is $1$. This latter fact is a result of Emma T. Lehmer; for a modern perspective (which hopefully contains a proof -- I haven't checked), see Gregory Dresden, Resultants of cyclotomic polynomials, Rocky Mountain Journal of Mathematics 42, Number 5, 2012.

$\endgroup$

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.