Show that the prime factors $p > 3$ of the integer $n^2 - n + 1$ are of the form 6k + 1.
Hint: Use the result of question 5 and let $p \mid (n^2− n + 1)$, and then consider $(2n− 1)^2$.
By the previous exercise, $ (2n-1)^2 \equiv -3 \pmod p $ requires any prime $p>3$ to conform to $p \equiv 1 \pmod 6$, that is $p=6k+1$ for some integer $k$.
We note that
$$ (2n-1)^2 = 4n^2 -4n +1 = 4(n^2 -n + 1) - 3 $$
This means such a $p$ divides $4(n^2 -n + 1)$. But $p \not \mid 4$ and so $p \mid (n^2 -n + 1)$.
And so we conclude the prime factors $p > 3$ of the integer $n^2 - n + 1$ are of the form 6k + 1.