 Solution

For odd coprime positive integers $a={\prod }_i{p}_i^{e_i}$ and $b={\prod }_j{q}_j^{f_j}$, define the Jacobi symbol by multiplying Legendre symbols over the prime factorization of the denominator. Multiplicativity in the numerator givesBy quadratic reciprocity for the Legendre symbol, this is $(-1{)}^{UV}$, where $U={\sum }_i{e}_i(p_i-1)/2$ and $V={\sum }_j{f}_j(q_j-1)/2$. For odd $x,y$, $(xy-1)/2\equiv (x-1)/2+(y-1)/2(\mathrm{mod}2)$, since $(x-1)(y-1)/2$ is even. Thus $U\equiv (a-1)/2$ and $V\equiv (b-1)/2(\mathrm{mod}2)$. We obtainIf $a,b$ are not coprime, both symbols are zero and the equivalent identity $(a/b)=(-1{)}^{(a-1)(b-1)/4}(b/a)$ still holds. The convention $(a/1)=1$ handles denominator one.