Problem regarding quadratic recidues that I cannot solve

Question

The problem is as follows:

Let $p$ be a prime and assume that $p=1$ mod 5. Let $c\in\left(\mathbb{Z}/p\mathbb{Z}\right)^{\times}$ be an element of order 5 and let $g=2(c+c^{-1})+1$ Show that $g^2=5$ mod $p$

Seeing as $g$ is given I thought it would be worthwile to compute $g^2$

I get that

$g^2 = (2(c+c^{-1})+1)^2=4(c+c^{-1})^2+4(c+c^{-1})+1$

$g^2=4(c^2+c^{-2}+2cc^{-1})+4(c+c^{-1})+1$

Then modulo $p$

$g^2=4(c^2+c^{-2}+2)+4(c+c^{-1})+1$ mod $p$

I do not know how to move on from here or if I am on the right track at all. I am not sure how to use the assumptions about $p$ and $c$. I can write $c^{-1}=c^4$ and $c^{-2}=c^8=c^3$ since c has order 5 but I'm not sure if that helps.

If anyone has any hints as to where to go, please let me know. It would be much appreciated.

Answer 1

Hint 1

An element of order $5$ is a root of $f = (x^{5} - 1)/(x - 1) = x^4 + x^3 + x^2 + x + 1$.

Hint 2

The roots of $f$ are $c, c^{2}, c^{3} = c^{-2}, c^{4} = c^{-1}$.

Hint 3

$c+ c^{2}+ c^{3}+ c^{4} = -1 =$ minus the coefficient of $x^{3}$ in $f$.