= Solution
For odd positive coprime integers, <quadratic reciprocity> for the <Jacobi symbol> is
$$
\boxed{\left(\frac mn\right)\left(\frac nm\right)=(-1)^{(m-1)(n-1)/4}.}
$$
If 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 $n=\prod_jq_j^{b_j}$. Multiplication of the prime reciprocity identities gives the sign exponent
$$
\sum_{i,j}a_ib_j\frac{p_i-1}{2}\frac{q_j-1}{2}
=\left(\sum_i a_i\frac{p_i-1}{2}\right)\left(\sum_j b_j\frac{q_j-1}{2}\right).
$$
Modulo two, each bracket is respectively $(m-1)/2$ and $(n-1)/2$: for odd $u,v$, $(uv-1)/2\equiv(u-1)/2+(v-1)/2\pmod2$. This proves the composite-denominator law, including repeated prime factors.
Now $261=9\cdot29$, and 9 contributes a square factor. Since $317\equiv1\pmod4$, reciprocity and reduction of the numerator give
$$
\left(\frac{261}{317}\right)=\left(\frac{29}{317}\right)=\left(\frac{27}{29}\right)=\left(\frac3{29}\right).
$$
A second reciprocity step gives $(3/29)=(29/3)(-1)^{14}=(2/3)=-1$. Thus \b[the requested <Jacobi symbol> is $-1$].
Back to article page
