Is $2^{218!} +1$ prime? [closed]
Prove that $2^{218!} +1$ is not a prime number.
I can prove that the last digit of this number is $7$, and that's all.
Thank you.
$$218!=3n:\;\;\;2^{218!}+1=(2^{n}+1)(4^{n}-2^{n}+1)$$
Well, $a^3+1=(a+1)(a^2-a+1)$, so it's not prime, and we have $$2^{218!}=(2^{2\cdot4\cdot5\cdot..\cdot 218})^3\,.$$
$x=1 \cdot 2 \cdot 4 \cdot 5 \dots 218$
$2^{218!}+1 =2^{3x}+1=(2^{x}+1)(2^{2x}-2^{x}+1)$
Another one: $2^{2^{213}}+1\mid2^{218!}+1$ because $218!=2^{213}\cdot k$ with $k$ odd.