Past exam of the options I course of the University of Oxford / 2008 / C / C1 / Solution

The quadratic reciprocity law states that distinct odd primes $a,b$ satisfyFirst suppose $p\ne q$. If the odd prime $r$ equalled one of them, divisibility of $p-q$ by $r$ would force the other to equal $r$ too, a contradiction. Thus reciprocity applies to $r,p$ and $r,q$. The modular congruence modulo $4r$ gives both $p\equiv q(\mathrm{mod}r)$ and $p\equiv q(\mathrm{mod}4)$. Therefore $(p/r)=(q/r)$, and the reciprocity signs are identical. Rearranging the two reciprocity identities proves $\boxed{(r/p)=(r/q)}$. If $p=q$, the equality is immediate, including a zero Legendre symbol when $p=r$.

For general odd $n>1$, factor $n={\prod }_j{r}_j^{e_j}$. If $p=q$ there is nothing to prove. Otherwise neither $p$ nor $q$ divides $n$, since any such divisor, together with $4n\mid p-q$, would force $p=q$. Each $4r_j$ divides $p-q$, so the first result and multiplicativity of the Legendre symbol giveFor $n=1$, both symbols equal one.