Finding non-trivial factor of $2^{40}+1$

Question

How is it possible to find a non-trivial factor of $2^{40}+1$?

I have no idea of which formula/procedure I should use. Can anybody help me?

https://math.stackexchange.com/questions/641443/proof-of-anbn-divisible-by-ab-when-n-is-odd — lab bhattacharjee, Apr 08 '19 at 15:48
Possible duplicate of Proof of $a^n+b^n$ divisible by $a+b$ when $n$ is odd — Jyrki Lahtonen, Apr 08 '19 at 17:50

score 4 · Answer 1 · answered Apr 08 '19 at 15:59

4

Observe that $$\begin{align} 2^{40}+1 & = (2^8)^5 + 1. \\ & = (2^8+1) ((2^8)^4-(2^8)^3+(2^8)^2-2^8+1). \end{align}$$

answered Apr 08 '19 at 15:59

little o

score 1 · Accepted Answer · answered Apr 08 '19 at 15:57

1

$40$ is divisible by $5$, so you can use the following factorization:

$2^{40}+1=(2^8+1)(2^{32}-2^{24}+2^{16}-2^8+1)$

since $a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^4+b^5)$

Therefore, a non trivial factor of $2^{40}+1$ is $2^8+1=257$

answered Apr 08 '19 at 15:57

user289143

BTW it turns out that $(2^{40}+1)/257$ is prime. – Robert Israel Apr 08 '19 at 16:23

2 Answers2