Prime divisors of $p^n+1$ Let $p$ be a rational prime and $n$ be a positive integer. 
It can be easily deduced from Zsigmondy's theorem that $p^n+1$ has a prime divisor greater than $2n$ except when $(p,n)=(2,3)$ or $(2^k-1,1)$ for some positive integer $k$. Hence we know that there exists an odd prime divisor of $p^n+1$ greater than $n$ if and only if $(p,n)\neq(2,3)$ or $(2^k-1,1)$ for any positive integer $k$.
Question: 
(1). For which $(p,n)$ does there exist at least two odd prime divisors of $p^n+1$ coprime to $n$?
(2).  For which $(p,n)$ does there exist at least two odd prime divisors of $p^n+1$ greater than $n$?

$x^n+1$ factors over $\mathbb{Z}[x]$ unless $n$ is a power of two.
For $n=15$ the factorization is $ (x + 1) \cdot (x^{2} - x + 1) \cdot (x^{4} - x^{3} + x^{2} - x + 1) \cdot (x^{8} + x^{7} - x^{5} - x^{4} - x^{3} + x + 1)$.
The factors $(p^{2} - p + 1)$ and $(p^{4} - p^{3} + p^{2} - p + 1)$ are odd and with congruence reasons you can get at least two prime factors larger than $15$.
