Solution to equation modulo p

Question

Under the assumptions that $$p\cong 1 \mod 5$$ and $$g = 2(c+c^{-1})+1$$ where $c$ has order $5$ modulo $p$. I need to show that $g^2 \cong 5 \mod p$. I have that $$g^2=4(c^4+c^3+c^2+c)+9$$ I know that if I can somehow remove the $c^3$ term, I can solve the equation through quadratic reciprocity. Can anybody give me some hints?

Answer 1

The key thing to note here is that $c$ has order $5$ modulo $p$; this implies that $c^5 \equiv 1 \pmod p$, and factoring gives $(c-1)(1+c+c^2+c^3+c^4) \equiv 0 \pmod p$. So either you will have $c=1$, which is handled trivially, or you will have $c+c^2+c^3+c^4 \equiv -1 \pmod p$, allowing you to simplify your expression.