TITLE: Can you solve current version of Quadratic reciprocity from Gauss version?
QUESTION [1 upvotes]: Can you solve current version( after: "He notes that these can be combined") of Quadratic reciprocity from Gauss version?
I have tried it but I have problem to understand especially if $ q \equiv 3 \pmod 4 \Rightarrow  (\frac{p}{q}) =1 \Leftrightarrow (\frac{-p}{q}) =1$.
I understand that if $ q \equiv 3 \pmod 4$(2) then by using Euler's criterion you get $(\frac{p}{q}) =1$, but don't know how you get  $(\frac{q}{p}) =1$(1). But is this(1) goal if you assume (2)?

REPLY [1 votes]: Gauss' version:
(1) Suppose $q\equiv 1\pmod{4}$.  Then $x^2\equiv p\pmod{q}$ is solvable if and only if $x^2\equiv q\pmod{p}$ is solvable.  Using Legendre symbols, this means exactly $\left(\frac{p}{q}\right)=\left(\frac{q}{p}\right)$.
(2) Suppose $q\equiv 3\pmod{4}$.  Then Then $x^2\equiv p\pmod{q}$ is solvable if and only if $x^2\equiv -q\pmod{p}$ is solvable.  Using Legendre symbols, this means exactly $\left(\frac{p}{q}\right)=\left(\frac{-q}{p}\right)$.
We may write $\left(\frac{-q}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{q}{p}\right)$.  But by the first supplement to QR, $\left(\frac{-1}{p}\right)=(-1)^{\frac{p-1}{2}}$. 
Hence, if $p\equiv 1\pmod{4}$, then $\left(\frac{p}{q}\right)=\left(\frac{q}{p}\right)$ or $\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=1=(-1)^{\frac{(p-1)(q-1)}{4}}$.
If instead $p\equiv 3\pmod{4}$, then $\left(\frac{p}{q}\right)=-\left(\frac{q}{p}\right)$ or $\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=-1=(-1)^{\frac{(p-1)(q-1)}{4}}$.
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