Show that $\sum_{j=0}^{2m} x^{2^nj} $ has at least $n+1$ distinct factors 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.
