Divisors of a cyclotomic polynomial I am not entirely sure of the validity of this source (https://proofwiki.org/wiki/Prime_Divisors_of_Cyclotomic_Polynomials), but it claims that all prime divisors $d$ of $\Phi_n(a)$ have $d\equiv 1\mod n$. 
I am able to prove that $gcd(d-1, n)>1$, but I am not able to reconcile the above statement to myself. For example, for $\Phi_{p^a}(a)$, are all prime factors still given by $d\equiv 1\mod p^a$. Or is it simply true that $d\equiv 1\mod p$? 
Can someone offer a straightforward proof of this? A paper or official reference of this?  
 A: I think your statement is wrong. For example, $\Phi_3(x)=x^2+x+1$, $\Phi_3(1)=3$ but $3\equiv 0\pmod 3$. However, we can prove the following statement (which is stated in your link).
Proposition. Suppose that for some positive integer n and integer $a$ there prime number $p$ such that $p|\Phi_n(a)$. Then, $p|n$ or $n|p-1$.
Proof. Consider case when $n$ is not divisible by $p$. We need to prove that $p|n-1$. Let $s$ be a minimal positive integer such that $p|a^s-1$. Assume that $s<n$ (it's obvious that $s\leq n$ because $p|a^n-1$). It's easy to see that $s|n$, so $n=ms$ for some positve integer $m$. Note that 
$$
\frac{a^{n}-1}{a^{s}-1}=\prod_{d|n,~d\nmid s}\Phi_d(a),
$$
so $\Phi_n(a)|\frac{a^{n}-1}{a^{s}-1}$ (at this point we use the inequality $s<n$). Since $p|\Phi_n(a)$ we obtain $p|\frac{a^{n}-1}{a^{s}-1}$. On the other hand,
$$
\frac{a^{n}-1}{a^{s}-1}=\frac{a^{ms}-1}{a^{s}-1}=a^{(m-1)s}+a^{(m-2)s}+\ldots+a^{s}+1\equiv m\pmod p
$$
because $a^s\equiv 1\pmod p$. Therefore, $p|m$, which impossible because $m|n$ and $p\nmid n$. We get a contradiction, so our assumption was incorrect. Hence, $s=n$. 
Using Fermat's Little Theorem we obtain $p|a^{p-1}-1$. But $n$ is minimal $k$ such that $p|a^k-1$, so $n|p-1$ and proposition is proved.
