Show that $\sum_{j=0}^{2m} x^{2^nj} $ has at least $n+1$ distinct factors.
This is a generalization (with $x=2$ and $m=1$) of this problem that I saw on quora:
Show that $2^{2^{n+1}}+2^{2^n}+1 $ has at least $n+1$ distinct divisors.
Here is my solution.
I use $x^{2^n}-1 =(x-1)\prod_{k=0}^{n-1}(x^{2^k}+1) $, $(x-1)\sum_{j=0}^{m-1} x^{j} =x^m-1 $ and $(x+1)\sum_{j=0}^{2m} (-1)^jx^{j} =x^{2m+1}+1 $.
Let $a_m(x) =\sum_{j=0}^{m-1} x^{j} $ and $b_m(x) =\sum_{j=0}^{m-1} (-1)^jx^{j} =a_m(-x) $. Then $(x-1)a_m(x) =x^m-1 $ and $(x+1)b_{2m+1}(x) =x^{2m+1}+1 $.
$\begin{array}\\ a_{2m}(x^{2^n}) &=\sum_{j=0}^{2m} x^{2^nj}\\ &=\dfrac{(x^{2^n})^{2m+1}-1}{x^{2^n}-1}\\ &=\dfrac{(x^{2m+1})^{2^n}-1}{x^{2^n}-1}\\ &=\dfrac{(x^{2m+1}-1)\prod_{k=0}^{n-1}((x^{2m+1})^{2^k}+1)}{(x-1)\prod_{k=0}^{n-1}(x^{2^k}+1)}\\ &=\dfrac{a_{2m}(x)\prod_{k=0}^{n-1}((x^{2^k})^{2m+1}+1)}{\prod_{k=0}^{n-1}(x^{2^k}+1)}\\ &=\dfrac{a_{2m}(x)\prod_{k=0}^{n-1}((x^{2^k}+1)b_{2m}(x^{2^k}))}{\prod_{k=0}^{n-1}(x^{2^k}+1)}\\ &=a_{2m}(x)\prod_{k=0}^{n-1}b_{2m}(x^{2^k})\\ \end{array} $
If $n>0$, then $$ \sum_{j=0}^{2m} x^{2^nj} = \frac{x^{2^n(2m+1)}-1}{x^{2^n}-1} = \left[\frac{x^{2^{n-1}(2m+1)}+1}{x^{2^{n-1}}+1}\right]\left[\frac{x^{2^{n-1}(2m+1)}-1}{x^{2^{n-1}}-1}\right] = \left[\sum_{j=0}^{2m} \left(-1\right)^j x^{2^{n-1}j}\right] \sum_{j=0}^{2m} x^{2^{n-1}j}. $$ By induction, $$ \sum_{j=0}^{2m} x^{2^nj} = \left(\sum_{j=0}^{2m}x^j\right)\prod_{k=0}^{n-1}\left[\sum_{j=0}^{2m} \left(-1\right)^j x^{2^{k}j}\right]. $$ I'm reasonably certain these polynomial factors are irreducible, so this gives the desired $n+1$ distinct factors.