Determine $\left ( \frac{5}{p} \right )=1$ in a different way

Question

Let $p$ prime with $p\equiv 1\pmod 5$. Let $c$ be an element of order $5$ in $(\mathbb{Z}/p\mathbb{Z})^\times$. I suppose this element exists, since the set $(\mathbb{Z}/p\mathbb{Z})^\times$ is cyclic. Putting $g=2\cdot (c+c^{-1})+1$, my lecturer wrote $g^2\equiv 5\pmod p$. It is left as an exercise, which I can't see how reach it. If I am not asking too much, how do I use it to show that $\left ( \frac{5}{p} \right )=1$?

Answer 1

Maybe this hint is enough: $c^{-1} = c^4$ and $c^{-2} = c^3.$

When you expand $g^2$ you can arrange it to be

$$4(1+c+c^2+c^3+c^4) + 5.$$

So it boils down to showing that $p$ divides $1+c+c^2+c^3+c^4.$