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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!

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  • $\begingroup$ 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. $\endgroup$ – Qiaochu Yuan Nov 17 '15 at 4:24
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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 $(2c+1)^2$.

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  • $\begingroup$ Thank you so much! That factorization really made it clear. $\endgroup$ – nightfjall Nov 17 '15 at 4:02

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