Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 1 / 17G (Quadratic Mathematics)

For a nonzero residue $a$ modulo an odd prime $p$, the Legendre symbol $(a/p)$ equals $+1$ if $a$ is a nonzero square modulo $p$ and $-1$ otherwise. The extension to multiples of $p$ has value zero. Euler's criterion states

Apply Euler's criterion to $-1$. It givesFor $p\nmid ab$, the same criterion gives $(ab/p)\equiv (ab{)}^{(p-1)/2}=a^{(p-1)/2}{b}^{(p-1)/2}\equiv (a/p)(b/p)(\mathrm{mod}p)$. Both sides are integers in $\{1,-1\}$, which are distinct modulo an odd prime, so congruence implies equality. This proves multiplicativity of the Legendre symbol:

Since $n$ is nonzero modulo $p$, it has a unique multiplicative inverse $m$ in the nonzero residues. The inverse cannot be $-1$ because that would imply $n=-1$, excluded by the range. Thus $1\le m\le p-2$ as claimed. The congruence $mn=1$ givesThe Legendre symbol of the nonzero square $n^2$ is one. Using multiplicativity of the Legendre symbol therefore gives

Inversion permutes the nonzero residues excluding $-1$. Hence the result of part (c) changes the requested sum intoThere are exactly $(p-1)/2$ nonzero squares: the map $x\mapsto x^2$ on nonzero residues identifies exactly the pairs $x,-x$, because $x^2=y^2$ implies $x=\pm y$ in the finite field. Thus there are equally many nonzero squares and nonsquares, and the complete Legendre-symbol sum is zero. Removing the $j=1$ term leaves