= 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 gives
$$
\chi(mn)\equiv (mn)^h=m^hn^h\equiv\chi(m)\chi(n)\pmod p.
$$
Both sides are $1$ or $-1$, which are distinct modulo the odd <prime number> $p$, so \b[$\chi(mn)=\chi(m)\chi(n)$.] Taking $n=-1$ in <Euler's criterion> gives \b[$\chi(-1)=(-1)^{(p-1)/2}$], which is $1$ for $p\equiv1\pmod4$ and $-1$ for $p\equiv3\pmod4$.
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