= Solution
If $p$ divides $a$ or $b$, both sides of the claimed <multiplicativity> identity are zero. Otherwise the <Euler criterion> gives
$$
\left(\frac{ab}p\right)\equiv (ab)^{(p-1)/2}
\equiv a^{(p-1)/2}b^{(p-1)/2}
\equiv\left(\frac ap\right)\left(\frac bp\right)\pmod p.
$$
Each side is $1$ or $-1$. Since $p$ is odd, these two integers represent different residue classes. Hence the <modular congruence> gives equality:
$$
\boxed{\left(\frac{ab}p\right)=\left(\frac ap\right)\left(\frac bp\right).}
$$
Back to article page
