Solution
= Solution
For odd $m,n$,
$$
\frac{mn-1}{2}=\frac{m-1}{2}+\frac{n-1}{2}+\frac{(m-1)(n-1)}2.
$$
The last term is even, so induction over the prime factors, including repetitions, gives $\frac{b-1}{2}\equiv\sum_i\frac{p_i-1}{2}\pmod2$. Using the first supplementary law in each Legendre factor,
$$
\boxed{\left(\frac{-1}b\right)=\prod_i(-1)^{(p_i-1)/2}=(-1)^{(b-1)/2}.}
$$
For $b=1$, the product is empty and both sides equal one.