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For $n\geq 2$, show that $$X^{\varphi(n)}\Phi_{n}(X^{-1})=\Phi_{n}(X)$$ where $\varphi$ is Euler totient function, $\Phi_{n}$ is the $n$th cyclotomic polynomial.

I've tried some discrete examples for some $n$, it all matches our result. And we have several formulas for the cyclotomic polynomial. But I don't know which one is most useful in our case?

[edited to correct statement]

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    $\begingroup$ Do you know the degree of the cyclotomic polynomial? Do you know that if $\zeta$ is a root of the cyclotomic polynomial then so is $\zeta^{-1}$? $\endgroup$ Commented Mar 30, 2017 at 9:56
  • $\begingroup$ @ancientmathematician I'm not very familiar with this property.Can you explain how the whole machinary is gonna work? $\endgroup$
    – Jack
    Commented Mar 30, 2017 at 11:31

1 Answer 1

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$\Phi_n(X)$ is the product of all $(X-\zeta)$ where $\zeta$ is an $n$-th root of unity but not a $d$-th root of unity for any proper divisor $d$ of $n$. Hence the degree of the $n$-th cyclotomic polynomial is $\varphi(n)$, the number of $k \in \left\{1,2,\ldots,n\right\}$ that are coprime with $n$.

Now $\zeta^k=1$ if and only if $(\zeta^{-1})^k=1$, so that $\zeta^{-1}$ is a root of $\Phi_n(X)$ whenever $\zeta$ is a root.

Now both $X^{\varphi(n)}\Phi_n(X^{-1})$ and $\Phi_n(X)$ are monic polynomials of the same degree with distinct roots, and every root $\zeta^{-1}$ of the first is a root of the second. Hence they are equal.

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  • $\begingroup$ Simple, beautiful, elegant. $\endgroup$
    – IAG
    Commented May 25 at 17:59

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