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Does there exist two cyclotomic polynomial $\Phi_n$ and $\Phi_m$ which are equal but $n\neq m$?

The cyclotomic polynomial is defined as $\Phi_n(x)=\prod_{\substack{1\le j\le n \\ \gcd(j,n)=1}}(x-u_{(j,n)})$ for which $u_{(j,n)}=e^{\frac{2\pi i j}{n}}$.

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Hitjuich
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    What is your definition of the cyclotomic polynomials? – Arthur Apr 24 '19 at 19:02
  • In a word, no, because the cyclotomic polynomials are pretty much defined by their roots, and the roots of $\Phi_n$ are disjoint from the roots of $\Phi_m$ when $n\neq m.$ Indeed, the polynomials are relatively prime when $n\neq m.$ – Thomas Andrews Apr 24 '19 at 19:04
  • It is possible to choose two numbers $m$ and $n$ such that $\phi(m)=\phi(n)$. Hence, they have the same numbers of factors. How can one be sure that the roots are disjoint? – Hitjuich Apr 24 '19 at 19:16
  • The roots of $\Phi_m$ are the primitive $m$'th roots of unity, i.e. $\omega$ is a root of $\Phi_m$ iff $\omega^m = 1$ but $\omega^k \ne 1$ for all $k < m$. – Robert Israel Apr 24 '19 at 19:40

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$\Phi_n$ and $\Phi_m$ are different for different $m$ and $n$, because they are coprime polynomials in $\Bbb Q[x]$ for $m\neq n$. Note that it is not enough to consider their degree, because we could have $\phi(m)=\phi(n)$ for different $m$ and $n$, e.g., $\phi(6)=\phi(3)=2$. However, $\Phi_6(x)=x^2-x+1$ and $\Phi_3(x)=x^2+x+1$.

Reference: When are cyclotomic polynomials coprime over the integers?

Dietrich Burde
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From your definition it is immediate that if $n\neq m$ then $$\Phi_n(u_{(1,n)})=0 \qquad\text{ and }\qquad \Phi_m(u_{(1,n)})\neq0,$$ which shows that $\Phi_n\neq\Phi_m$.

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