Past exam of the options II for mathematics and philosophy course of the University of Oxford / 2013 / C2

For an odd prime $p$, the Legendre symbol is zero if $p\mid a$, one if $a$ is a nonzero quadratic residue modulo $p$, and minus one if it is a quadratic nonresidue. ThusIt depends only on the congruence class of $a$ modulo $p$.

The first supplementary law for quadratic reciprocity isIt also follows immediately from Euler's criterion applied to $-1$.

The second supplementary law for quadratic reciprocity isThe exponent is an integer because every odd square is one modulo eight.

Multiplicativity of the Legendre symbol givesThis includes the case where $p$ divides a factor, when both sides are zero.

For distinct odd primes $p,q$, the quadratic reciprocity law isThus the two symbols have opposite signs exactly when both primes are three modulo four; otherwise they agree. Distinctness excludes zero symbols.

Factor $a={\prod }_i{q}_i$ and $b={\prod }_j{p}_j$, listing prime factors with repetitions. Coprimality ensures $q_i\ne p_j$. By the definition of the Jacobi symbol and multiplicativity of the Legendre symbol,For odd integers $r,s$,since the difference is $(r-1)(s-1)/2$, an even integer. Inductively the two sums in the exponent are congruent to $(a-1)/2$ and $(b-1)/2$ modulo two. HenceIf either number is one, the empty product convention proves the same identity. This extends the prime law to arbitrary coprime positive odd denominators.

Put $p=4272943$. Since $p\equiv 7(\mathrm{mod}8)$, the supplementary laws give $(-1/p)=-1$ and $(2/p)=1$. Therefore $(-286/p)=-(143/p)$.

Both $143$ and $p$ are three modulo four. Reciprocity for the Jacobi symbol, followed by the supplied remainder, givesAgain both denominators are three modulo four, soNow $103\equiv 7(\mathrm{mod}8)$ and $40=8\cdot 5$. Consequentlybecause the nonzero squares modulo five are one and four. Thus $(103/143)=1$, $(143/p)=-1$, and finallyThe congruence is soluble, with exactly two distinct solutions modulo this prime: a nonzero root and its negative. The computation uses a prime denominator in its final step; a positive Jacobi symbol with composite denominator would not by itself imply solubility.