How to prove that $\sum_{i=0}^kx^{in}$ have a same factor $P(x)$ if and only if $n \not\equiv 0 \bmod (k+1)$？ Please see the pre-content here:
How to prove that $x^{2n}+x^n+1$ has a factor $x^2+x+1$ if and only if $n \not\equiv 0 \bmod 3$？
Well, I don't know whether it is exactly correct. I get this result ($x^{2n}+x^n+1$ has a factor $x^2+x+1$ if and only if $n \not\equiv 0 \bmod 3$) when I was dealing with a simple problem "find out the factors of $x^{10}+x^5+1$". I put different $n$ into this polynomial and draw the roots in the complex plane by WolframAlpha:
n=4,7,9
The distribution of these roots has certain rules: the red angle is half of the blue one, and one red angle comes after one blue angle and make up the whole circle in this way. By observing the pictures, we can easily get the result that is stated in the question.
Also, for $x^{3n}+x^{2n}+x^n+1$, I infer it has a factor $x^2+1$ if and only if $n\not \equiv 0\bmod 4$. Also see the pictures: n=2,3,4,5
My guess is:
For polynomials $x^{kn}+x^{(k-1)n}+…+x^{2n}+x^n+1,k\geq2$, if $n\not \equiv 0\bmod (k+1)$, they all have a certain factor $P(x)$.
Is that true？How do I find the factor $P(x)$？How to prove？(I think the method of Cotes' Theorem may be used for it. See T. Needham, Visual Complex Analysis, pp 46-47)
 A: If $P_k(x)=P(x)=x^{k}+x^{(k-1)}+…+x^{2}+x+1$, then with $Q_{k,n}(x)=Q(x)=x^{kn}+x^{(k-1)n}+…+x^{2n}+x^n+1$ we have that $Q(y)=P(y^n)$.
Since the roots of $P$ are precisely the $k+1$ roots of unity distinct from $1$, so $e^{2\pi im/(k+1)}, m=1,..k$ we get that the roots of $Q$ are precisely $e^{2\pi ir/n}e^{2\pi im/(n(k+1))}, m=1,..k, r=0,..n-1$
Now if $(n, k+1)=1$ one can easily see that one can solve in $c=1,..k, r=0,..n-1$ the equation $r(k+1)-cn=m$ for every $m=1,..k$ as one first solves $r(k+1)-cn=1$ with $(c, k+1)=1$ and then one gets the solutions for $m=2,..k$ as $2c,...kc$ (reducing them modulo $k+1$) which are none divisible by $k+1$ so are indeed in $1,..k$ modulo $k+1$
But this means that for all such $n$ we have that $P_k/Q_{k,n}$ so indeed all $Q_{k,n}$ have that as a common subfactor.
If $(n,k+1)=d >1$ then we get as roots of $Q$  only the nontrivial roots of unity of order $(k+1)$ (and of course roots of unity of higher order involving $n$) that are of the form $e^{2\pi idm/(k+1)}, m=1,..(k+1)/d$  so in general, there is no common factor.
