From the book,
Suppose $p \equiv 1 \pmod{4}$, then by law of quadratic reciprocity, we have: $$\left(\frac{3}{p}\right) = \left(\frac{p}{3}\right) $$ Next, if $p \equiv 2 \pmod{3}$, then $p \equiv 5 \pmod{12}$ Hence, $$\left(\frac{3}{p}\right) = \left(\frac{p}{3}\right) = -1$$
How do they get those Legendre fraction equal to $-1$? From my understanding: $$\left(\frac{q}{p}\right) \cdot \left(\frac{p}{q}\right) = (-1)^{\frac{p-1}{2}\cdot\frac{q-1}{2}}$$ So $q = 3 \implies \frac{q-1}{2} = \frac{3-1}{2} = 1$ For $p$, I take $p \equiv 5 \pmod{12} \implies p = 5 + 12k$, for some integers k.
Hence, $\frac{p-1}{2} = \frac{12k + 5 - 1}{2} = \frac{12k + 4}{2} = 6k + 2.$ And this $6k + 2$ is even :( ! How does $(-1)^{even} = -1$?
Any idea? I think I made some logic mistakes somewhere, but I couldn't find where.
Update
The problem was from Elementary Number Theory and Its Application - Kenneth H.Rosen 5th Edition.
Problem
Using the law of quadratic reciprocity, show that if $p$ is an odd prime, then $$\left(\frac{3}{p}\right) = 1 \text{ if } p \equiv \pm 1 \pmod{12}$$ $$\left(\frac{3}{p}\right) = -1 \text{ if } p \equiv \pm 5 \pmod{12}$$Solution
Thanks,
Now I'm even more confused :(!
Consider two cases:
Case 1 $$p \equiv 1 \pmod{4} \text{ and } p \equiv 1 \pmod{3}$$ Then,
$$p \equiv 1 \pmod{12} \implies p = 12k + 1$$
Hence,
$$\frac{p - 1}{2} \cdot \frac{3 - 1}{2} = \frac{12k}{2} = 6k = \text{ even }$$
Which implies
$$\left(\frac{3}{p}\right) \cdot \left(\frac{p}{3}\right) = 1$$
So both must be $1$ or $-1$.
Case 2 $$p \equiv 1 \pmod{4} \text{ and } p \equiv 2 \pmod{3}$$ Then,
$$p \equiv 5 \pmod{12} \implies p = 12k + 5$$
Hence,
$$\frac{p - 1}{2} \cdot \frac{3 - 1}{2} = \frac{12k + 4}{2} = 6k + 2 = 2(3k + 1) = \text{ even }$$ Which implies
$$\left(\frac{3}{p}\right) \cdot \left(\frac{p}{3}\right) = 1$$ So both must be $1$ or $-1$.
I don't see how these arguments can be deduced to the solution. Any suggestion?