Past exam of the short options course of the University of Oxford / 2021 / 1 / a

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.

Factor $Q={\prod }_i{p}_i$ and $Q^{\prime }={\prod }_j{q}_j$, retaining repeated prime factors. If $P$ has a common prime factor with $Q{Q}^{\prime }$, it has one with at least one of $Q,Q^{\prime }$, and both sides of the desired equality are zero. Otherwise the definition of the Jacobi symbol givesAn empty product is $1$, so the argument includes either denominator being $1$.

For a prime number $p$, multiplicativity of the Legendre symbol says $(P{P}^{\prime }/p)=(P/p)(P^{\prime }/p)$, including the case of a zero factor. Therefore, writing $Q={\prod }_i{p}_i$ with repetitions,This product definition still agrees with the stipulated zero when the numerator and denominator are not coprime. For $Q=1$ all three symbols are $1$.

If $P\equiv P^{\prime }(\mathrm{mod}Q)$, then $P\equiv P^{\prime }(\mathrm{mod}p)$ for each prime factor $p$ of $Q$. The Legendre symbol depends only on the numerator's residue class, so the corresponding factors agree. Also $\gcd (P,Q)=\gcd (P^{\prime },Q)$, covering the zero cases. Multiplying givesThe convention for denominator $1$ gives the same conclusion there.