 Solution

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 $p\equiv 3(\mathrm{mod}4)$.