Exercise 5.38

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Let $p$ be an odd prime. Derive the quadratic character of $2$ modulo $p$ by verifying the following steps, involving the Jacobi symbol:

$$
 \legendre{2}{p} = \legendre{8-p}{p} = \legendre{p}{p-8} = \legendre{8}{p-8} = \legendre{2}{p-8}.
$$

Generalize the argument to show that
$$
  \left(\frac{a}{p}ight) = \left(\frac{a}{p-4a}ight), \qquad a>0, p 
mid a.
$$

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

(As in Ex. 5.36, since $8 - p$ or $p-8$ is negative, we interpret $(a/b)$ as the Kronecker symbol : see definition in Ex. 5.36.)
\begin{proof}
As $(2^2/p) = 1$ and $8-p \equiv 8 \pmod p$,
$$ \legendre{2}{p} = \legendre{2^2}{p} \legendre{2}{p} = \legendre{8}{p} = \legendre{8-p}{p}.$$
As $p$ and $8-p$ are odd numbers and $p>0$, from the extension of the law of quadratic reciprocity to $a<0$ proved in Ex. 5.36, we obtain
$$\legendre{8-p}{p} = (-1)^{\frac{7-p}{2} \frac{p-1}{2}} \legendre{p}{8-p}.$$
Moreover
\begin{align*}
(7-p)(p-1) &\equiv (-1-p)(p-1) = 1 - p^2 \pmod 8\
\end{align*}
As $p= 2k+1$ is odd, $p^2 = 4k^2+4k+1 = 8 \frac{k(k+1)}{2} + 1 \equiv 1 \pmod 8$, so $(7-p)(p-1) \equiv 0 \pmod 8$ and $\frac{7-p}{2} \frac{p-1}{2}$ is even, so
$$\legendre{8-p}{p} = \legendre{p}{8-p}.$$
As $p>0$, $\legendre{p}{-1} = 1$, thus $ \legendre{p}{8-p} = \legendre{p}{-1}  \legendre{p}{p-8} = \legendre{p}{p-8}$ (with the same argument, this is also true for the  3 odd primes such that $8-p>0$), so
$$\legendre{8-p}{p} = \legendre{p}{p-8}.$$
\end{proof}
As $p\equiv 8 \pmod{p-8}$, $\legendre{p}{p-8}  = \legendre{8}{p-8}$, and since $8 = 2^2	imes 2$, $\legendre{8}{p-8} = \legendre{2}{p-8}$. We have proved for all odd primes $p$ that
$$  \legendre{2}{p} = \legendre{8-p}{p} = \legendre{p}{p-8} = \legendre{8}{p-8} = \legendre{2}{p-8}.$$

The preceding arguments remain valid if we replace the odd prime $p$ by any odd positive integer.  So with an immediate induction, we see that for all $k \in \N$,
$$ \legendre{2}{p} = \legendre{2}{p-8k}.$$
So the quadratic character of $2$ modulo $p$ depends only of the class of $p$ modulo $8$.

If $p \equiv 1 \pmod 8$, $\legendre{2}{p} =\legendre{2}{1} = 1$.

If $p \equiv -1 \pmod 8$, $\legendre{2}{p} =\legendre{2}{-1} = 1$.

If $p \equiv \pm 3 \pmod 8$, $\legendre{2}{p} =\legendre{2}{\pm 3} = -1$.

\bigskip

Generalization: let $a>0$ and $p$ be an odd positive integer such that $p \wedge a =1$ (not necessarily prime).

$$\legendre{a}{p} = \legendre{4a p}{p} = \legendre{4a-p}{p} = (-1)^{\frac{4a - p -1}{2} \frac{p-1}{2}} \legendre{p}{4a-p}.$$
$(4a-p-1)(p-1)=  4a(p-1)  + 1 - p^2 \equiv 0 \pmod 8$, so 
$$\legendre{a}{p} =  \legendre{p}{4a-p}.$$
As $\legendre{p}{-1} = 1$,
$$  \legendre{p}{4a-p} =   \legendre{p}{p-4a}.$$
Since $p \equiv 4a \pmod {p-4a}$,  and $4$ is a square,
$$\legendre{p}{p-4a} \equiv \legendre{4a}{p-4a} =  \legendre{a}{p-4a}.$$

We have proved
$$\legendre{a}{p} = \legendre{4a-p}{p} =  \legendre{p}{p-4a} = \legendre{4a}{p-4a}  =  \legendre{a}{p-4a}.$$

By induction, for all $k\geq 0$, $\legendre{a}{p} = \legendre{a}{p-4ka}$, so $\legendre{a}{p}$ depends only of the class of $p$ modulo $4a$.

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Exercise 5.38

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Exercise 5.38

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