# An equality about cyclotomic polynomials

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]

• 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 $\zeta^{-1}$? Mar 30 '17 at 9:56
• @ancientmathematician I'm not very familiar with this property.Can you explain how the whole machinary is gonna work?
– mike
Mar 30 '17 at 11:31

$$\Phi_n(X)$$ 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.