Exercise 3.2.11 (Is $x^2 \equiv a \pmod {pq}$ solvable, where $a$ is a quadratic nonresidue mod $p$ and mod $q$)

Proof. The conclusion is never true. Suppose that $a$ is a quadratic nonresidue of each of the odd primes $p$ and $q$. Assume for contradiction that $x^2\equiv a(\mathit{mod}\mathit{pq})$ for some integer $x$. Since $p\mid \mathit{pq}$, $x^2\equiv a(\mathit{mod}p)$. Therefore $p\mid a$, or $a$ is a quadratic residue modulo $p$. But both are false, since $a$ is a quadratic nonresidue modulo $p$. Therefore $x^2\equiv a(\mathit{mod}\mathit{pq})$ is not solvable.

A counterexample is given by $a=3,p=5,q=7$. Then $3$ is a non quadratic nonresidue for $5$ and $7$, and $x^2\equiv 3(\mathit{mod}35)$ is not solvable (otherwise $x^2\equiv 3(\mathit{mod}5)$ would be solvable). □