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

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.