Jacobi reciprocity law ($(a/b)(b/a)=(-1{)}^{(a-1)(b-1)/4}$)

For coprime positive odd integers $a,b$, the Jacobi symbol satisfies $(a/b)(b/a)=(-1{)}^{(a-1)(b-1)/4}$. It follows by factoring $a$ and $b$ into primes and multiplying the quadratic reciprocity for each prime pair; the parity of the sums of prime half-differences agrees with $(a-1)/2$ and $(b-1)/2$. A Jacobi symbol equal to one does not imply that its numerator is a quadratic residue for a composite denominator.

For an odd positive nonsquare integer $a$, there are infinitely many primes $p$ with Legendre symbol $(a/p)=-1$. Choose a residue $n$ modulo $a$ with Jacobi symbol $(n/a)=-1$ by making it a quadratic nonresidue at one prime occurring to odd multiplicity and a square at every other prime. The Chinese remainder theorem also imposes $n\equiv 1(\mathrm{mod}4)$ and $n\equiv 1$ at any finite list of previously found primes. Then Jacobi reciprocity gives $(a/n)=-1$, so a prime factor of $n$ has $(a/p)=-1$ and is outside that list.