Exercise 9.2.13

Proof. By Exercise 8 (b), we have proved for $p=17$, that

So

Let

$\phi$ is a group homomorphism.

As $x^2=1⟺(x-1)(x+1)=0⟺x\in \{-1,1\}$, $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\{-1,1\}\subset {(\mathbb{Z}∕\mathit{p\mathbb{Z}})}^{\text{∗}}$. Write $C=\mathrm{im}(\phi )$ the set of square elements in ${(\mathbb{Z}∕\mathit{p\mathbb{Z}})}^{\text{∗}}$. Then $\mathrm{im}(\phi )\simeq {(\mathbb{Z}∕\mathit{p\mathbb{Z}})}^{\text{∗}}∕\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$, so $|C|=|\mathrm{im}(\phi )|=(p-1)∕2=8$. Moreover $H_8=⟨3^2⟩$ (Exercise 1), so $H_8\subset C$, and $|H_8|=8=|C|$, therefore $H_8=C$ is the set of squares in ${(\mathbb{Z}∕17\mathbb{Z})}^{\text{∗}}$. Its complement $3H_8$ is the set of non squares in ${(\mathbb{Z}∕17\mathbb{Z})}^{\text{∗}}$.

Therefore, for all $a\in {(\mathbb{Z}∕17\mathbb{Z})}^{\text{∗}}$.

and $\left(\frac{a}{17}\right)=0$ if $a=0$ or $a=17$ (where we write for all integer $k$, $\left(\frac{[k]}{17}\right)=\left(\frac{k}{17}\right))$. Hence

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