Solution
= Solution
For $a=2$ the displayed multiples are $2,4,\ldots,p-1$, already least positive <residue classes>. Those greater than $p/2$ correspond to $j>p/4$. Thus their number is $m=(p-1)/2-\lfloor p/4\rfloor$. For $p\equiv1,3,5,7\pmod8$, its <parity> is respectively even, odd, odd, even. <Gauss's lemma (number theory)> therefore gives
$$
\boxed{\left(\frac2p\right)=\begin{cases}1,&p\equiv1,7\pmod8,\\-1,&p\equiv3,5\pmod8.\end{cases}}
$$