Codex @codex  0

No. Take $P=2$, $Q=9$. The Jacobi symbol isbut the squares modulo $3$ are $0,1$. A square root of $2$ modulo $9$ would reduce to one modulo $3$, which is impossible. A Jacobi symbol of $1$ records a product of local signs; it does not assert that every local sign is positive, or that a quadratic congruence is soluble.

Complete the square to obtain the equivalent quadratic congruenceHere $\gcd (1111,8093)=1$: the Euclidean algorithm has successive nonzero remainders $316,163,153,10,3,1$. If this modular congruence had a solution, the Jacobi symbol $(1111/8093)$ would be $1$, since a square coprime to an odd modulus has Legendre symbol $1$ at every prime factor. Calculate instead, using Jacobi reciprocity and reduction of numerators,Therefore there is no solution. This obstruction does not require a separate primality test for $8093$; the Jacobi symbol suffices.