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Let $p>3$ be a prime number. Is it true that $$ \left(\frac{-3}{p}\right)= \begin{cases} 1 & p\equiv1(\bmod{\,3})\\ -1 & p\equiv-1(\bmod{\,3})\\ \end{cases}\quad? $$ Thanks!

boaz
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2 Answers2

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If $p\equiv 1\pmod{3}$, by Cauchy's theorem there is an order-$3$ element of $\mathbb{Z}/(p\mathbb{Z})^*$.
If we call such element $\omega$, we have $\omega^3\equiv 1\pmod{p}$ and $\omega\not\equiv 1\pmod{p}$, hence $$ \omega^2+\omega+1 \equiv 0\pmod{p} \tag{1}$$ $$ 4\omega^2+4\omega+1 \equiv -3\pmod{p} \tag{2}$$ $$ (2\omega+1)^2 \equiv -3\pmod{p}\tag{3} $$ and $-3$ is a quadratic residue. Conversely, if $-3$ is a quadratic residue for some prime $p>3$, there must be an order-$3$ element of $\mathbb{Z}/(p\mathbb{Z})^*$, hence $3\mid(p-1)$ by Lagrange's theorem and $p\equiv 1\pmod{3}$.

Jack D'Aurizio
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  • Which way are you deducing that an order $3$ element exists if $-3$ is a quadratic residue? – Bill Dubuque May 28 '17 at 21:15
  • @BillDubuque: by following the inverse path $(3)\mapsto(2)\mapsto(1)$. If $a^2\equiv -3$, then $\frac{a-1}{2}$ is a root of $x^2+x+1=0$, hence an order-$3$ element. – Jack D'Aurizio May 28 '17 at 21:21
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Using quadratic reciprocity $$ \begin{align} \left(\frac{-3}{p}\right)&=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right) \\ &=\left(\frac{-1}{p}\right)\left(\frac{-1}{p}\right)\left(\frac{p}{3}\right) \\ &= \left(\frac{p}{3}\right) \end{align} $$ And it's easy to check that squares are $\equiv 1\bmod{3}$.

sharding4
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  • That is correct but I think that the interesting part comes from proving such result without resorting to the full power of quadratic reciprocity. Indeed, Cauchy and Lagrange's theorems do the job pretty nicely. – Jack D'Aurizio May 28 '17 at 17:19
  • Agreed and your post strikes a nice balance between being complete and not overcomplicating things. – sharding4 May 28 '17 at 17:25