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Question. Show $$\Big(\frac{-3}{p}\Big)=\begin{cases}+1 & p\equiv 1\bmod 3,\\ -1 & p\equiv 2\bmod 3. \end{cases}$$

Attempt. So, using the established results of $$\Big(\frac{3}{p} \Big) = \begin{cases}+1 & p\equiv\pm1\bmod 12,\\ -1 & p\equiv\pm5\bmod 12,\end{cases}$$ $$\Big(\frac{-1}{p}\Big)=\begin{cases}+1 & p\equiv1\bmod 4,\\ -1 & p\equiv3\bmod 4,\end{cases}$$ and observing that $$\Big(\frac{-1}{p} \Big) = \begin{cases}+1 & p\equiv1,5\bmod 12,\\ -1 & p\equiv7,11\bmod 12,\end{cases}$$ we get $$\Big(\frac{-3}{p}\Big)=\begin{cases}+1 & p\equiv 1,7\bmod 12,\\ -1 & p\equiv 5,11\bmod 12. \end{cases}$$ Providing I've carried this out right; how can I reduce this further so it is in the form of modulo $3$?

Michael Hardy
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thesmallprint
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2 Answers2

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You have that $p\equiv 1,7\pmod{12}$ if and only if $p\equiv 1\pmod{3}$ and $p$ odd.

Similarly for $p\equiv 5,11\pmod{12}$ and $p$ odd with $p\equiv 2\pmod{3}.$\

There is an easier approach if you know quadratic reciprocity:

$$\begin{align}\left(\frac{-3}{p}\right)&=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)\\ &=\left(-1\right)^{\frac{p-1}{2}}\left(-1\right)^{\frac{p-1}{2}\frac{3-1}{2}}\left(\frac{p}{3}\right)\\ &=(-1)^{p-1}\left(\frac p3\right)\\&=\left(\frac{p}{3}\right) \end{align}$$

More generally, when $q\equiv 3\pmod{4}$ is a prime you get that:

$$\left(\frac{-q}{p}\right)=\left(\frac{p}{q}\right)$$

Thomas Andrews
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$p \equiv 1 \bmod 3$ iff $p \equiv 1,4,7,10 \bmod 12$, but $p \not\equiv 4,10 \bmod 12$ because $p$ is odd.

$p \equiv 2 \bmod 3$ iff $p \equiv 2,5,8,11 \bmod 12$, but $p \not\equiv 2,8 \bmod 12$ because $p$ is odd.

lhf
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