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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?

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closed as off-topic by Saad, Xander Henderson, user296602, Jyrki Lahtonen, A. Pongrácz Apr 8 at 18:15

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$40$ 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 $2^8+1=257$

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  • $\begingroup$ BTW it turns out that $(2^{40}+1)/257$ is prime. $\endgroup$ – Robert Israel Apr 8 at 16:23
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Observe that $$\begin{align} 2^{40}+1 & = (2^8)^5 + 1. \\ & = (2^8+1) ((2^8)^4-(2^8)^3+(2^8)^2-2^8+1). \end{align}$$

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