Show that the quadratic reciprocity theorem could also be written (p/q)(q/p) = (-1)^[(p-1)(q-1)/4] for odd primes p and q.
Here are the four cases:
$\displaystyle p\equiv1\mod{4} $ and $\displaystyle q\equiv1\mod{4} $
$\displaystyle p\equiv1\mod{4} $ and $\displaystyle q\equiv3\mod{4} $
$\displaystyle p\equiv3\mod{4} $ and $\displaystyle q\equiv1\mod{4} $
$\displaystyle p\equiv3\mod{4} $ and $\displaystyle q\equiv3\mod{4} $
I'll verify one case for you:
$\displaystyle p\equiv1\mod{4} $ and $\displaystyle q\equiv3\mod{4} $
We know $\displaystyle \left(\frac{p}{q}\right) = \left(\frac{q}{p}\right) $.
Therefore $\displaystyle \left(\frac{p}{q}\right)\left(\frac{q}{p}\right) = \left(\frac{p}{q}\right)^2 = (\pm1)^2 = 1 $
We're also given that $\displaystyle p=4k+1 $ and $\displaystyle q=4n+3 $.
So $\displaystyle \tfrac14(p-1)(q-1) = 2k(2n+1) = 2x $.
Thus $\displaystyle (-1)^{\frac{(p-1)(q-1)}{4}} = (-1)^{2x} = 1 $.
(Notice that is proves the case $\displaystyle p\equiv3\mod{4} $ and $\displaystyle q\equiv1\mod{4} $ too since $\displaystyle p $ and $\displaystyle q $ are arbitrary.