Let $p>3$. Suppose $f$ is third primitive root of unity in $F_p$. I need to prove the following: $(2f+1)^2=(-3)$ .
So, what I did: $f^3=1$ but it doesn't hold for first and second power of $f$. Then, $(2f+1)^2 = 4(f^2 + f + 1) - 3$
But then, if I multiply the LHS by $f$, I will get $f^3 + f^2 + f = 1 + f^2 + f$ . So, $f$ must be $1$. I am clearly confusing something here. I am doing this too straightforward?