Let $a = p_1^{k_1} \times p_2^{k_2} \times \ldots \times p_n^{k_n}$ be the prime decomposition of $a$. Let $p$ be prime such that $p \not \mid a$. Prove that
$$ (\frac{a}{p}) = (\frac{p_1}{p})^{k_1} \times (\frac{p_2}{p})^{k_2} \times \ldots \times (\frac{p_n}{p})^{k_n} $$
where $(\frac{m}{p})$ is the Legendre symbol.
We proceed as follows
$$ \begin{align} (\frac{a}{p}) & = (\frac{p_1^{k_1} \times p_2^{k_2} \times \ldots \times p_n^{k_n}}{p}) \\ \\ &= (\frac{p_1^{k_1}}{p}) \times (\frac{p_2^{k_2}}{p}) \times \ldots \times (\frac{p_n^{k_n}}{p}) \\ \\ &= \biggl (\underbrace{(\frac{p_1}{p}) \times (\frac{p_1}{p}) \ldots (\frac{p_1}{p})}_{k_1} \biggr ) \times \biggl ( \underbrace{(\frac{p_2}{p}) \times (\frac{p_2}{p}) \ldots (\frac{p_2}{p})}_{k_2} \biggr ) \times \ldots \times \biggl (\underbrace{(\frac{p_1}{p}) \times (\frac{p_n}{p}) \ldots (\frac{p_n}{p})}_{k_n} \biggr ) \\ \\ &= (\frac{p_1}{p})^{k_1} \times (\frac{p_2}{p})^{k_2} \times \ldots \times (\frac{p_n}{p})^{k_n} \end{align} $$
The second and third steps use the result from the previous exercise.