$p|\Phi_n(2)$ then $p|\Phi_{pn}(2)$

Question

Prove that $p|\Phi_n(2)$ then $p|\Phi_{pn}(2)$

Here $\Phi_n(x)$ is nth cyclotomic polynomial.

I don't know what I should use. $$\Phi_n(x)=\prod_{{1\leq a\leq n } \& {(a,n)=1}}(x-\zeta_n^a) $$ or $$\Phi_n(x)=\prod_{d|n}(x^{n/d}-1)^{\mu(d)}$$

Answer 1

Observe that the map $(-)^p:\mathbb F_p[x]\to \mathbb F_p[x]$ is ring morphism (with trivial kernel). For any positive integer $m$ indivisible by prime $p$ and nonnegative integer $k$ one has $$\begin{align*}\Phi_{p^km}(x)&=_{\mathbb F_p[x]}\prod_{d\mid p^km}\left(x^{p^km/d}-1\right)^{\mu(d)}\\ &=_{\mathbb F_p[x]}\prod_{d\mid m}\left(\left(x^{p^km/d}-1\right)\left(x^{p^{k-1}m/d}-1\right)^{-1}\right)^{\mu(d)}\\ &=_{\mathbb F_p[x]}\prod_{d\mid m}\left(\left(x^{m/d}-1\right)^{p^k-p^{k-1}}\right)^{\mu(d)}\\ &=_{\mathbb F_p[x]}\left(\prod_{d\mid m} (x^{m/d}-1)^{\mu(d)}\right)^{p^k-p^{k-1}}=_{\mathbb F_p[x]} \Phi_{m}(x)^{p^k-p^{k-1}}\end{align*}$$ From which it follows that $$\Phi_n(2)=_{\mathbb F_p}0\implies \Phi_{n/\gcd(n,p^\infty)}(2)=_{\mathbb F_p}0\implies \Phi_{pn}(2)=_{\mathbb F_p}0$$ Where $\gcd(n,p^\infty)$ is a shorthand for the largest power of $p$ dividing $n$ $\blacksquare$