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