Question 9 of 9.3 Elementary number theory David Burton

Question

I am trying section 9.3 of David Burton Elementary Number Theory and got struck on this problem.

If $p$ and $q$ are odd primes satisfying $p=q+4a$ for some a establish that $\big(\frac{a}{p}\big)$ = $\big(\frac{a}{q}\big)$.

Hint is given: Note that $\big(\frac{a}{p}\big) =\big(\frac{-q}{p}\big)$ and use the Quadratic Reciprocity Law.

Attempt:I proved that $\big(\frac{a}{p}\big) =\big(\frac{-q}{p}\big)$ and tried using reciprocity law on $\big(\frac{-q}{p}\big)$ but unable to proceed.

Can you please tell if I am doing it right and please tell right way.

Thanks a lot!!

Also this is not a homework problem. I am self studying. You can answer after as many days as you please.

Answer 1

$$\bigg(\frac{a}{p}\bigg)=\bigg(\frac{-q}{p}\bigg)=\bigg(\frac{-1}{p}\bigg)\cdot\bigg(\frac{q}{p}\bigg)=\bigg(\frac{-1}{p}\bigg)\cdot\frac{(-1)^{\frac{(p-1)(q-1)}{4}}}{\bigg(\frac{p}{q}\bigg)}=(-1)^{\frac{p-1}{2}}\cdot\frac{(-1)^{\frac{(p-1)(q-1)}{4}}}{\bigg(\frac{4a}{q}\bigg)}=(-1)^{\frac{p-1}{2}}\cdot\frac{(-1)^{\frac{(p-1)(q-1)}{4}}}{\bigg(\frac{a}{q}\bigg)}$$

So we have $$\bigg(\frac{a}{p}\bigg)=(-1)^{\frac{p-1}{2}}\cdot\frac{(-1)^{\frac{(p-1)(q-1)}{4}}}{\bigg(\frac{a}{q}\bigg)}$$ so

$$\bigg(\frac{a}{p}\bigg)\cdot\bigg(\frac{a}{q}\bigg)=(-1)^{\frac{2p-2+(p-1)(q-1)}{4}}=(-1)^{\frac{(p-1)(q+1)}{4}}$$

As $p\equiv q\pmod{4}$, then $\frac{(p-1)(q+1)}{4}$ is even so $$\bigg(\frac{a}{p}\bigg)\cdot\bigg(\frac{a}{q}\bigg)=1$$ so $$\bigg(\frac{a}{p}\bigg)=\bigg(\frac{a}{q}\bigg)$$

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EDIT: Here is how I got it, Ben:

$p\equiv q\pmod{4}$ means that either both $p$ and $q$ are $\equiv 1\pmod{4}$ or $\equiv 3\pmod{4}$.

Suppose $p,q\equiv 3\pmod{4}$. Then $p-1\equiv 2\pmod{4}$ and $q+1\equiv 0\pmod{4}$ so $(p-1)(q+1)$ is divisible by $8$, thus $\frac{(p-1)(q+1)}{4}$ is even.

Similarly, if $p,q\equiv 1\pmod{4}$ then $p-1\equiv 0\pmod{4}$ and $q+1\equiv 2\pmod{4}$ so again $(p-1)(q+1)$ is divisible by $8$, thus $\frac{(p-1)(q+1)}{4}$ is even.

Answer 2

As you noted, we have $(a/p)=(-q/p)$, and similarly, we have $(a/q)=(p/q)$, hence

\begin{align*} & (a/p)=(a/q) \\[4pt] \iff & (-q/p)=(p/q) \\[4pt] \iff & (-1/p)(q/p)=(p/q) \\[4pt] \iff & (-1/p)(q/p)^2=(p/q)(q/p) \\[4pt] \iff & (-1/p)=(p/q)(q/p) \\[4pt] \iff & (-1)^{{\Large{\frac{p-1}{2}}}}=(-1)^{{\Large{\frac{p-1}{2}}}{\Large{\frac{q-1}{2}}}} \\[4pt] \iff & \frac{p-1}{2}\equiv \frac{p-1}{2}{\,\cdot\,}\frac{q-1}{2}\;(\text{mod}\;2) \\[4pt] \iff & \frac{p-1}{2}{\,\cdot\,}\frac{q-1}{2}-\frac{p-1}{2} \equiv 0\;(\text{mod}\;2) \\[4pt] \iff & \left(\frac{p-1}{2}\right)\left(\frac{q-1}{2}-1\right) \equiv 0\;(\text{mod}\;2) \\[4pt] \iff & \frac{p-1}{2}{\,\cdot\,}\frac{q-3}{2} \equiv 0\;(\text{mod}\;2) \\[4pt] \end{align*} Then from $p-q=4a$, we get $p\equiv q\;(\text{mod}\;4)$, hence

• If $p\equiv q\equiv 1\;(\text{mod}\;4)$, then ${\large{\frac{p-1}{2}}}$ is even.$\\[4pt]$
• If $p\equiv q\equiv 3\;(\text{mod}\;4)$, then ${\large{\frac{q-3}{2}}}$ is even.

so in either case we get $$ \frac{p-1}{2}{\,\cdot\,}\frac{q-3}{2} \equiv 0\;(\text{mod}\;2) \qquad $$ which completes the proof.