# Use that $(\mathbb{Z}/p\mathbb{Z})^{*}$ is cyclic to give a direct proof that $\left( \frac{-3}{p} \right) = 1$ when $p \equiv 1\ (\bmod\ 3)$.

Use that $(\mathbb{Z}/p\mathbb{Z})^{*}$ is cyclic to give a direct proof that $\left( \frac{-3}{p} \right) = 1$ when $p \equiv 1\ (\bmod\ 3)$.

(Hint: There is an element $c \in (\mathbb{Z}/p\mathbb{Z})^{*}$ of order 3. Show that $(2c+1)^2 = -3$.)

I could really use some help with some direction on where to start with this proof. I've gone through a some examples and I can see that it works, but I can't seem to prove it for the general case.

Any responses would be much appreciated!

• You don't need to know that $\mathbb{Z}_p^{\times}$ is cyclic to know that it contains an element of order $3$; you can just appeal to Lagrange's theorem. – Qiaochu Yuan Nov 17 '15 at 4:24

Hint: Since $c$ is a third root of unity, $c$ is a root of $x^3-1$ in $\mathbb F_p = \mathbb Z/p\mathbb Z$. Moreover, $c \neq 1$, so $c$ is a root of the polynomial obtained by factoring $x-1$ out of $x^3-1$, which is $x^2+x+1$. (Note that you can only do this factoring because $\mathbb F_p$ is an integral domain.)
Use the fact that $c^2+c+1=0$ to simplify $(2c+1)^2$.