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.)
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.