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automation studio 5.2 crack full versionQ:

Existence of a whole number $k$ such that $n^2+n+1$ is divisible by $k$

Prove there exists a whole number $k$ such that $n^2+n+1$ is divisible by $k$.

I was thinking that I should use the fact that $n^2+n+1=n(n+1)+1$ can be written as $(n+1)^2+n+1$ and $(n+1)^2$ can be written as $n(n+1)$.
Note:
I don’t know what a “whole number” means. I think it means that it can’t be integer not-prime numbers.
Thanks.

A:

$n^2+n+1 = (n+1)^2+1$. Since $2$ is coprime to $n+1$, $n+1$ is an integer $\equiv 0,1,2,3,\ldots\pmod 2$, and thus an odd number. Hence there is an odd whole number $m$ satisfying $m\mid n^2+n+1\Leftrightarrow 2m\mid (n+1)^2+n+1\Leftrightarrow 2m\mid (n+1)^2+2$.
But $2m\mid (n+1)^2+2$ is impossible, since $(n+1)^2+2$ is even.
So the only option is $2m\mid (n+1)^2+1$, i.e. $m\mid (n+1)^2+1$. But since $(n+1)^2+1$ is odd, $m$ is even. So we have $m\mid (n+1)^2+1$.
Now we want to prove there is some integer $m$ such that $m\mid n+1$.
By Euclid’s lemma, there is a primitive root of $m$ (i.e. it’s the smallest integer $r$ satisfying $r\mid m\implies r^k\mid m$ for some \$k http://southfloridafashionacademy.com/wp-content/uploads/2022/06/Butterfly_Abstract_Windows_7_Theme.pdf

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