Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 2 / 1G (Number Theory)

For odd positive coprime integers, quadratic reciprocity for the Jacobi symbol isIf the integers are not coprime, both symbols vanish; equivalently the usual signed reciprocity identity still holds. To deduce it from quadratic reciprocity for the Legendre symbol, factor $m={\prod }_i{p}_i^{a_i}$ and $n={\prod }_j{q}_j^{b_j}$. Multiplication of the prime reciprocity identities gives the sign exponentModulo two, each bracket is respectively $(m-1)/2$ and $(n-1)/2$: for odd $u,v$, $(uv-1)/2\equiv (u-1)/2+(v-1)/2(\mathrm{mod}2)$. This proves the composite-denominator law, including repeated prime factors.

Now $261=9\cdot 29$, and 9 contributes a square factor. Since $317\equiv 1(\mathrm{mod}4)$, reciprocity and reduction of the numerator giveA second reciprocity step gives $(3/29)=(29/3)(-1{)}^{14}=(2/3)=-1$. Thus the requested Jacobi symbol is $-1$.