# Finding non-trivial factor of $2^{40}+1$ [closed]

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?

## closed as off-topic by Saad, Xander Henderson, user296602, Jyrki Lahtonen, A. PongráczApr 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 $$2^8+1=257$$

• BTW it turns out that $(2^{40}+1)/257$ is prime. – Robert Israel Apr 8 at 16:23

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}