Past exam of the options II course of the University of Oxford / 2013 / C2 / c / Solution

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.