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.
 A: 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 $\phi(1)=1,\phi(3)=2,\phi(5)=4,\phi(15)=8$.
A: 
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 $1366\times768$. :-)
