Past exam of the Probability, statistics and options II course of the University of Oxford / 2011 / C / C2 / c

Because $a=-b^2$ is odd, $b$ is odd. Choose $k=4b^2-1$, a positive integer congruent to three modulo four. Also $\gcd (k,b)=1$, so the square $b^2$ has Jacobi symbol one modulo $k$. Multiplicativity in the numerator follows from the corresponding Legendre property in each prime factor. ThusThis gives an explicit positive odd denominator with the required obstruction.

Choose a nonzero quadratic nonresidue $c$ modulo $q$. By the Chinese remainder theorem, choose a positive integer $k$ satisfyingThe moduli are pairwise coprime, and $k$ is coprime to $qb$. Jacobi reciprocity, with $k\equiv 1(\mathrm{mod}4)$, givesSince $t$ is odd, multiplying these symbols yieldsWhen $b=1$ its congruence imposes no restriction and its Jacobi factor equals one.

An integer square is a square modulo every prime, with the zero class allowed at primes dividing the integer. Conversely, suppose an odd integer $a$ is not an integer square. If every prime exponent in $|a|$ is even, the nonsquare condition forces $a=-b^2$, handled by part (i). Otherwise some odd prime has an odd exponent, so $a=\pm q^{t}b$ as in part (ii). In either case there is a positive odd $k$ with $(a/k)=-1$.

Factor $k$ into odd primes. A product of Legendre symbols equalling minus one has at least one factor equal to minus one. Thus $a$ fails to be a square modulo one of those primes, contradicting the hypothesis. HenceThe solvability formulation makes explicit the zero-residue convention needed by the printed final assertion. The prime two causes no extra issue because every odd integer is one modulo two.