Factoring $x^n + y^n$ over the integers

Here's what i know (or think i know) about the factoring.

For integer $n> 1 $

1) If $n$ is a positive power of $2$ then it is irreducible.

2) If $n$ is an odd prime then $$x^n + y^n = (x + y)(x^{n-1} - x^{n-2}y + \cdots - xy^{n-2} + y^{n-1} ) $$

3) If $n$ has an odd prime factor then it is factorable but the factorization is more complicated , for example $x^{14} + y^{14}$ has 2 distinct irreducible factors and $x^{15} + y^{15}$ has $4$ distinct irreducible factors.

Is there a connection between the prime factorization of $n$ and the number of distinct irreducible factors of $x^n + y^n$ ?

Is there a connection between $n$, AND the number of distinct irreducible factors , AND the highest power occurring in each factor? Example:

$$x^{15} + y^{15} = (x + y)(x^2 - \cdots)(x^4 - \cdots)(x^8 + \cdots)$$

In other words , i'm also asking if there is a connection between $n = 15$ , the number of factors $4$ , and the powers $\{1 , 2 , 4 , 8\}$

For this particular example the numbers work out nicely but i'm not sure the pattern is so obvious in general.

Thank you for your consideration in this matter.

First, see Way to show $x^n + y^n = z^n$ factorises as $(x + y)(x + \zeta y) \cdots (x + \zeta^{n-1}y) = z^n$.

Since we have this factorisation, we need to know in what minimal way we need to combine the $n$th roots of unity so that we have a polynomial over the integers again. Well, if we multiply together all the primitive $d$th roots of unity terms for all $d\mid n$, that's what we want. There are $\phi(d)$ primitive $d$th roots of unity.

Hence you will have $\#\{1\leq d<n:d\mid n\}$ irreducible factors of degrees $\phi(d)$.

In your case, we have $1,3,5,15\mid 15$, and $\phi(1)=1,\phi(3)=2,\phi(5)=4,\phi(15)=8$.

Is there a connection between $n=15$, the number of factors $4$, and the powers $\{1,2,4,8\}$

For $n=2^{2^k}-1=\displaystyle\sum_{j=0}^{2^k-1}2^j$ , we have $\dfrac{x^n+y^n}{x+y}=\dfrac{x^{2^{2^k}-1}+y^{2^{2^k}-1}}{x+y}=\displaystyle\prod_{j=1}^{2^k-1}\Big(x^{2^j}\pm\ldots\Big)$. Here, $k=2$, and $n=15$. The cases $k<4$ are less than a few lines in length, while $k=4$ occupies about $40$ screens of resolution $1366\times768$. :-)