Quadratic character of 3

Question

Using the QRL prove that, for any odd prime $p$, $(3/p) = 1$ if $p$ is congruent to $1$ or $11 \pmod{12}$.

Using the Quadratic reciprocity law, $(3/p)(p/3)=(-1)^{(3-1)(p-1)/4}$, I get that the Legendre symbol is always equal to $1$.

What am I missing?

Because if p is odd the p-1 is even and therefore 2(even)=even which divided by 4 gives me an even power, therefore 1. — Christian, Dec 12 '13 at 20:51
No @Christian, what you get is that $;\frac{(3-1)(p-1)}2;$ is an integer, but not necessarily an even one ! — DonAntonio, Dec 12 '13 at 20:53
Next, find the primes $p \equiv 1 \pmod 3$ for which $3$ is a cube. That is $t^3 - 3 \equiv 0 \pmod p$ has a solution. — Will Jagy, Dec 12 '13 at 21:21

score 1 · Answer 1 · answered Dec 12 '13 at 20:50

From quadratic reciprocity, you get $$\left(\frac{3}{p}\right)\left(\frac{p}{3}\right) = (-1)^\frac{(3-1)(p-1)}{4} = (-1)^\frac{p-1}{2}$$ For us to get $\left(\frac{3}{p}\right) = 1$, we must have $$\left(\frac{p}{3}\right) = (-1)^\frac{p-1}{2}$$ If both are equal to $1$, then from $(-1)^\frac{p-1}{2}$ we get $4\mid p-1$ which implies $p\equiv 1 \pmod 4$. We must also have $p\equiv 1 \pmod 3$ since $\left(\frac{p}{3}\right)=1$. From the chinese remainder theorem, this happens if and only if $p\equiv 1 \pmod{12}$.

What if both are equal to $-1$? Can you finish off the argument?

DonAntonio · Answer 2 · 2013-12-12T20:58:06.717

1

Note that

$$p=\begin{cases}\;\;1\\11\end{cases}\pmod{12}\implies p=\begin{cases}1\\11\end{cases}+12k\;,\;\;k\in\Bbb N$$

so that

$$(3-1)(p-1)=2\left(\begin{cases}0\\10\end{cases}+12k\right)=\begin{cases}0\\20\end{cases}+24k=\begin{cases}24k\\20+24k\end{cases}$$

and thus

$$(-1)^{\frac{(3-1)(p-1)}2}=(-1)^{\begin{cases}12k\\10+12k\end{cases}}=1$$

edited Dec 12 '13 at 20:58

answered Dec 12 '13 at 20:52

DonAntonio

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Quadratic character of 3

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