- $\Phi_n(x^r) = \Phi_{nr}(x)$ where $r$ is a positive integer and if prime number $p|r$ then $p|n$.
- $\Phi_n(x^{ps}) = \Phi_{n}(x^s)\Phi_{pn}(x^s)$ where $p$ is a prime number, $p\nmid n$ and $s$ is a positive integer.
- $\Phi_n(x) = \Phi_{p_1 \dots p_k}(x^{np_1^{-1} \dots p_k^{-1}})$ where n is a positive integer and $p_1,\dots, p_k$ are different prime factors of $n$.
First I tried to prove the first identity. Using the definition of cyclotomic polynomial, we have $$\Phi_{nr}(x) = \prod_{1\le k \le nr\\ (k,nr)=1} (x-e^{\frac{2k\pi}{nr}\ i})$$ and $$\Phi_n(x^r) = \prod_{1\le k \le n \\ (k,n)=1} (x^r-e^{\frac{2k\pi}{n}\ i})$$
We know that when $1\le k \le n$, $(k,n)=1 \Leftrightarrow (k,nr)=1$.
How can I proceed from here?
Or is there any reference regarding these identities?
Here are some ideas. With all of these we start from
$$\Phi_n(x) = \prod_{d|n} (x^d-1)^{\mu(n/d)}.$$
For the first one we get with the prime factors of $r$ being a subset of those of $n$ that
$$\Phi_{rn}(x) = \prod_{d|rn} (x^{d}-1)^{\mu(rn/d)} = \prod_{d|rn} (x^{rn/d}-1)^{\mu(d)}.$$
Now as $r$ does not contribute any extra prime factors and the Mobius function is restricting terms this becomes
$$\prod_{d|n} (x^{rn/d}-1)^{\mu(d)} = \Phi_n(x^r).$$
For the second one we get with $p$ a prime not dividing $n$ that
$$\Phi_n(x^{ps}) = \prod_{d|n} (x^{psd}-1)^{\mu(n/d)} = \prod_{d|pn \wedge p|d } (x^{sd}-1)^{\mu(n/(d/p))} \\ = \prod_{d|pn \wedge p|d } (x^{sd}-1)^{\mu(pn/d)} = \prod_{d|pn} (x^{sd}-1)^{\mu(pn/d)} \prod_{d|n} (x^{sd}-1)^{-\mu(pn/d)} \\ = \Phi_{pn}(x^s) \prod_{d|n} (x^{sd}-1)^{\mu(n/d)} = \Phi_{pn}(x^s) \Phi_{n}(x^s).$$
For the third one we again have
$$\Phi_n(x) = \prod_{d|n} (x^{n/d}-1)^{\mu(d)}.$$
With $\mu(d)$ zero when $n$ is not squarefree this becomes
$$\prod_{d|p_1\cdots p_k} (x^{n/d}-1)^{\mu(d)}.$$
Now in this last product $n/d$ can be obtained by decrementing the power of zero, one or more of the prime factors of $n$ by one. In other words it is obtained by multiplying $n/p_1\cdots p_k$ by a divisor $f$ of $p_1\cdots p_k.$ We then find $d$ by solving $n/d = (n/p_1\cdots p_k) f$ to get $d = p_1\cdots p_k/f.$ Substitute these into the formula to obtain
$$\prod_{f|p_1\cdots p_k} (x^{(n/p_1\cdots p_k) f}-1)^{\mu(p_1\cdots p_k/f)}.$$
This is $\Phi_{p_1\cdots p_k}(x^{n/p_1\cdots p_k})$ by inspection.