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Let $m > 1$ and let $p$ be a prime not dividing $m$. If $\Phi_*$ denotes that $*$th cyclotomic polynomial, then establish the following identity:

$$\Phi_{pm}(x) = \frac{\Phi_m(x^p)}{\Phi_m(x)}$$

I have some intuition why this ought to hold. In particular, both sides should have the same roots (counting multiplicities of course), but I am having difficulty showing this. Could anyone help fill in the details?

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Using the identity $$\Phi_m(x) = \prod_{d|m} (x^d-1)^{\mu(m/d)}$$ we get for the RHS that $$\frac{\Phi_m(x^p)}{\Phi_m(x)} = \prod_{d|m} \frac{(x^{pd}-1)^{\mu(m/d)}}{(x^d-1)^{\mu(m/d)}}.$$ For the LHS we get by classifying divisors according to whether they are multiples of $p$ or not that $$\Phi_{pm}(x) = \prod_{d|m} (x^d-1)^{\mu(pm/d)} \prod_{d|m} (x^{pd}-1)^{\mu(pm/p/d)}.$$ But now $\mu(mp/d)=-\mu(m/d)$ since $p$ is prime and does not divide $m/d$ and we are done.

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