I'm trying to simplify $2^{80} - 1$ but having problems with factorizing $2^x+1$ elements.
What I got:
$2^{80}-1 = (2^{40})^2 - 1 = (2^{40} -1)(2^{40}+1) = ((2^{20})^2 - 1)(2^{40}+1) = (2^{20}-1)(2^{20}+1)(2^{40} + 1) = (2^{10} - 1)(2^{10} + 1)(2^{20}+1)(2^{40} + 1) = (2^5-1)(2^5+1)(2^{10} + 1)(2^{20}+1)(2^{40} + 1)$
So the problem is how do I factorize the $2^x+1$ elements, is there a formula which works for some cases?
Hint: Use the general formula $$a^k+b^k = (a+b)(a^{k-1}-a^{k-2}b + a^{k-3}b^2 -\ldots - ab^{k-2}+b^{k-1})$$ for $a,b \in \mathbb{R}$ and $k\in \mathbb{N}$ odd.
For instance, $2^{40}+1 = (2^8)^5+1^5 = 256^5+1^5 = 257 \cdot (256^4 - 256^3 + 256^2 - 256 +1)$