= Solution
If $\gcd(b,c)>1$, both <Jacobi symbol>[Jacobi symbols] vanish, so the formula holds. Otherwise write $b=\prod_i p_i$ and $c=\prod_j q_j$, retaining repeated factors. <Multiplicativity of the Jacobi symbol> and <quadratic reciprocity> give
$$
\left(\frac bc\right)
=\prod_{i,j}\left(\frac{p_i}{q_j}\right)
=\prod_{i,j}\left(\frac{q_j}{p_i}\right)
(-1)^{\sum_{i,j}(p_i-1)(q_j-1)/4}.
$$
The exponent is
$$
\left(\sum_i\frac{p_i-1}2\right)
\left(\sum_j\frac{q_j-1}2\right)
\equiv\frac{b-1}2\frac{c-1}2\pmod2
$$
by the parity identity from part (i). The product of the reversed <Legendre symbol>[Legendre symbols] is $(c/b)$. Thus the <Jacobi reciprocity law> is
$$
\boxed{\left(\frac bc\right)=\left(\frac cb\right)(-1)^{(b-1)(c-1)/4}.}
$$
This also covers $b=1$ or $c=1$ by empty products.
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