How many positive integers values of n there such that $2^{2^{2020}}-1 $ divisible by $ 2^{2^{n}}+1$. I found think a piece of full solution is :
let, $2^{(ab)}-1=(2^a-1)(1+2+2^2+..2^{((a-1)b))}$
Here $ab=2^{2020}.$
We can take $ a=2^n$ for $n≤2020.$ for such an even$ a , 2^n -1$ is divisible by$ (2^n/2)+1$
So for all $x ≤1010 $;
$2^x+1$ will divide $2^{(2020)}-1$
How to do the full solution.
there are $2019$ such values. Clearly $n>=2020$ won't work.
We prove that $2^{2^i}-1$ divides $2^{2^{2020}}-1$ for all $2\le i \le 2020$
the proof is by descent and boils down to $2^{2^i}-1=(2^{2^{i-1}}+1)(2^{2^{i-1}}-1)$
We can also use the expression $2^{2^i}-1=(2^{2^{i-1}}+1)(2^{2^{i-1}}-1)$
to notice $2^{2^i}+1$ divides $2^{2^{2020}}-1$ for all $1\leq i \leq 2019$