= Solution
If $b,c$ have a common prime factor, both Jacobi symbols are zero, and the identity holds. Otherwise write $b=\prod_i p_i$, $c=\prod_jq_j$ with multiplicity. All cross-pairs of primes are distinct. Multiplying <quadratic reciprocity> over these pairs gives
$$
\left(\frac bc\right)=\left(\frac cb\right)
(-1)^{\sum_{i,j}[(p_i-1)/2][(q_j-1)/2]}.
$$
The exponent factors as a product of two sums. The parity identity from part (i) makes it congruent modulo two to $[(b-1)/2][(c-1)/2]$. Therefore
$$
\boxed{\left(\frac bc\right)=\left(\frac cb\right)(-1)^{(b-1)(c-1)/4}.}
$$
The same formula includes a denominator equal to one by the empty-product convention.
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