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?
Best Answer
I believe you can find the value of $(3|p)$ without finding $p\equiv 5\pmod{12}$, by use of the second supplement to quadratic reciprocity. Recall that $$ \left(\frac{2}{p}\right)=(-1)^{(p^2-1)/8}. $$ Since $p\equiv 2\pmod{3}$, you then have $(p|3)=(2|3)=-1$. So altogether, $$ \left(\frac{3}{p}\right)\left(\frac{p}{3}\right)=\left(\frac{3}{p}\right)\left(\frac{2}{3}\right)=\left(\frac{3}{p}\right)(-1)=(-1)^{(p-1)(3-1)/4}=(-1)^{(p-1)/2}=1 $$ This implies $(3|p)=-1$.