Proof.
[First Proof of Theorem
4.1.7]
First suppose that
![$ p\equiv q\pmod{4}$](img1354.png)
. By swapping
![$ p$](img14.png)
and
![$ q$](img15.png)
if
necessary, we may assume that
![$ p>q$](img1104.png)
, and write
![$ p-q=4a$](img1355.png)
. Since
![$ p=4a+q$](img1356.png)
,
and
Proposition
4.3.4 implies that
![$ \left(\frac{a}{q}\right) = \left(\frac{a}{p}\right)$](img1359.png)
, since
![$ p\equiv q\pmod{4a}$](img1360.png)
. Thus
where the last equality
is because
![$ \frac{p-1}{2}$](img1362.png)
is even if and only if
![$ \frac{q-1}{2}$](img1363.png)
is
even.
Next suppose that
, so
.
Write
. We have
![$\displaystyle \left(\frac{p}{q}\right) = \left(\frac{4a-q}{q}\right) = \left(\frac{a}{q}\right),$](img1367.png)
and
Since
![$ p\equiv -q\pmod{4a}$](img1369.png)
,
Proposition
4.3.4 implies that
![$ \left(\frac{a}{q}\right) = \left(\frac{a}{p}\right)$](img1359.png)
. Since
![$ (-1)^{\frac{p-1}{2}\cdot
\frac{q-1}{2}}=1$](img1370.png)
, the proof is complete.