Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 4 / 11I (Number Theory)

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.

Factor $n={\prod }_j{q}_j^{e_j}$. Since it is not a square, one exponent, say $e_1$, is odd. Choose a nonzero quadratic nonresidue $a$ modulo $q_1$; one exists because the squaring map on the nonzero residues is two-to-one and hence has only $(q_1-1)/2$ images. The Chinese remainder theorem supplies a positive odd integer $m>1$ satisfyingThen $\gcd (m,n)=1$ and $(m/n)=(-1{)}^{e_1}=-1$. The Jacobi reciprocity law proved in part (i), together with $m\equiv 1(\mathrm{mod}4)$, gives $(n/m)=-1$. Factoring $m={\prod }_{\ell }{p}_{\ell }^{r_{\ell }}$ givesAll $p_{\ell }$ are odd and none divides $n$. At least one factor must be negative, yielding an odd prime number $\boxed{p\text{ with }(n/p)=-1}$. No theorem about primes in arithmetic progressions is needed.