Prove Corollary (7.17).
We remind ourselves of Prove Corollary (7.17).
Let p and q be distinct odd primes, then
$$ (\frac{p}{q}) = \begin{cases} (\frac{q}{p}) & \text{ if } p \equiv 1 \pmod 4 \text{ or } q \equiv 1 \pmod {4} \\ -(\frac{q}{p}) & \text{ if } p \equiv q \equiv 3 \pmod {4} \end{cases} $$
We use the result we proved in Exercise (7.4).3. If $p$ and $q$ are distinct odd primes then
$$ (\frac{p}{q}) \times (\frac{q}{p}) = \begin{cases} 1 & \text{ if } p \equiv 1 \pmod {4} \text{ or } q \equiv 1 \pmod {4} \\ -1 & \text{ if } p \equiv q \equiv 3 \pmod {4} \end{cases}$$
Let's consider each of the two cases in turn.
If $p \equiv 1 \pmod 4$ or $q \equiv 1 \pmod 4$ then
$$ (\frac{p}{q}) \times (\frac{q}{p}) = 1$$
This is only possible if $ (\frac{p}{q})$ and $(\frac{q}{p})$ are both 1 or both -1, that is $(\frac{p}{q}) = (\frac{q}{p})$. The corollary holds for this case.
If $p \equiv q \equiv 3 \pmod 4$ then
$$ (\frac{p}{q}) \times (\frac{q}{p}) = -1$$
This is only possible $ (\frac{p}{q}) = 1$ and $(\frac{q}{p})=-1$, or $ (\frac{p}{q}) = -1$ and $(\frac{q}{p})=1$. That is $(\frac{p}{q}) = -(\frac{q}{p})$. The corollary holds for this case.
We have shown Corollary (7.17) holds for all possible cases of $p$ and $q$.