= Solution
{c}
Write $P=\prod_i p_i$, $Q=\prod_j q_j$, counting repetitions. Since $P,Q$ are <coprime>, every pair $p_i,q_j$ consists of distinct odd <prime numbers>. <Quadratic reciprocity> for the <Legendre symbol> gives
$$
\left(\frac{p_i}{q_j}\right)=(-1)^{((p_i-1)/2)((q_j-1)/2)}\left(\frac{q_j}{p_i}\right).
$$
Multiplying over all $i,j$ and using <multiplicativity of the Jacobi symbol> yields the sign exponent
$$
\left(\sum_i\frac{p_i-1}{2}\right)\left(\sum_j\frac{q_j-1}{2}\right)\equiv\frac{P-1}{2}\frac{Q-1}{2}\pmod2,
$$
by the odd-product parity identity proved in (b)(i). Thus the <Jacobi reciprocity law> is
$$
\boxed{\left(\frac P Q\right)=(-1)^{(P-1)(Q-1)/4}\left(\frac Q P\right).}
$$
If either integer is $1$, its factorization is empty and both sides equal $1$.
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