Quadratic Residues and the Quadratic Equations

Question

Prove that, if $n$ is an integer, then $n^2 + 2n + 6$ is not divisible by 2017.

I know that 2017 is prime. I also have a feeling that I may be able to use the quadratic formula to help me. I just don't know how!

Answer 1

(Without using the rules for quadratic reciprocity:)

$n^2 + 2n + 6 = (n+1)^2+5$

If $2017$ divides $(n+1)^2+5$, this would mean $(n+1)^2\equiv -5 \bmod 2017$.

Since $2017$ is prime, by Fermat's Little Theorem, $(n+1)^{2016}\equiv 1 \bmod 2017$. Thus we can check $-5^{2016/2}=-5^{1008}=5^{1008}$ to see if that number can be a square. IF $5^{1008}\equiv 1\bmod 2017$ then we will be able to find an $n$ for the given formula, otherwise not.

Using exponentiation by squaring, we can find that $-5^{1008}\equiv 5^{1008}\equiv 2016\bmod 2017$, so there is no $n$ that fulfills $n^2 + 2n + 6 \equiv 0\bmod 2017$