= Solution
Factor $a=\prod_i q_i$ and $b=\prod_j p_j$, listing prime factors with repetitions. Coprimality ensures $q_i\ne p_j$. By the definition of the <Jacobi symbol> and <multiplicativity of the Legendre symbol>,
$$
(a/b)(b/a)=\prod_{i,j}(q_i/p_j)(p_j/q_i)
=(-1)^{\left(\sum_i(q_i-1)/2\right)\left(\sum_j(p_j-1)/2\right)}.
$$
For odd integers $r,s$,
$$
\frac{rs-1}{2}\equiv\frac{r-1}{2}+\frac{s-1}{2}\pmod2,
$$
since the difference is $(r-1)(s-1)/2$, an even integer. Inductively the two sums in the exponent are congruent to $(a-1)/2$ and $(b-1)/2$ modulo two. Hence
$$
\boxed{(a/b)(b/a)=(-1)^{\frac{a-1}{2}\frac{b-1}{2}}.}
$$
If either number is one, the empty product convention proves the same identity. This extends the prime law to arbitrary coprime positive odd denominators.
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