Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 2 / 1G (Number Theory)

For an odd prime number $p$ and an integer $n$ not divisible by $p$, the Legendre symbol $\chi (n)=(n/p)$ is $1$ if $n$ is a quadratic residue modulo $p$, and $-1$ otherwise. One may extend it by $\chi (n)=0$ when $p\mid n$.

Put $h=(p-1)/2$. The nonzero quadratic residues in the finite field ${\mathbb{F}}_p$ form a subgroup of size $h$: the group homomorphism $x\mapsto x^2$ has kernel $\{1,-1\}$. Each square satisfies $u^h=1$ by Fermat's little theorem. Since a nonzero polynomial of degree $h$ has at most $h$ roots of a polynomial, these are all the roots of $X^h-1$. For any other nonzero $u$, Fermat's theorem gives $(u^h{)}^2=1$, so $u^h=-1$. Thus Euler's criterion follows, and multiplication givesBoth sides are $1$ or $-1$, which are distinct modulo the odd prime number $p$, so $\chi (mn)=\chi (m)\chi (n)$. Taking $n=-1$ in Euler's criterion gives $\chi (-1)=(-1{)}^{(p-1)/2}$, which is $1$ for $p\equiv 1(\mathrm{mod}4)$ and $-1$ for $p\equiv 3(\mathrm{mod}4)$.

Euler's criterion gives $\chi (-1)=(-1{)}^{(p-1)/2}=1$ in this case. There are equally many nonzero quadratic residues and quadratic nonresidues, so ${\sum }_{n=1}^{p-1}\chi (n)=0$. Replacing $n$ by $p-n$ and using the multiplicativity of the Legendre symbol givesConsequently $\boxed{S=0}$.