Exercise 4.19

Proof.

(a)

If $p=7$, then $3\mid p-1,d=3\wedge (p-1)=3$. From Prop. 4.2.1, $$\exists <!--\; FUNCTION\; APPLICATION\; -->x\in \mathbb{Z},a\equiv x^3(\mathit{mod}7)⟺a\equiv 0(mod7)\mathrm{or}{a}^{(p-1)∕3}=a^2\equiv 1(mod7).$$

So the numbers $a$ such that $x^3\equiv a(\mathit{mod}7)$ is solvable are congruent at $0,1,-1$ modulo $7$.

(b)

If $p=11$, then $d=3\wedge (p-1)=1$. With the same proposition, $$\exists <!--\; FUNCTION\; APPLICATION\; -->x\in \mathbb{Z},a\equiv x^3(\mathit{mod}11)⟺a\equiv 0(mod11)\mathrm{or}{a}^{p-1}=a^6\equiv 1(mod11).$$

So all integers $a$ are cube modulo $11$, in only one way.

For an alternative proof, the application

$$f:\{\begin{array}{ccc}\hfill {𝔽}_{11}^{\text{∗}}\hfill & \hfill \text{→}\hfill & \hfill {𝔽}_{11}^{\text{∗}}\hfill \\ \hfill x\hfill & \hfill \mapsto \hfill & \hfill x^3\hfill \end{array}$$

$f$ is a bijection. Indeed,

$\text{∙}$ $f$ is a group homomorphism,

$\text{∙}$ $x^3=1\Rightarrow {(x^3)}^7=1\Rightarrow {(x^{10})}^{2}x=1\Rightarrow x=1$ thus $\ker <!--\; FUNCTION\; APPLICATION\; -->(f)=\{1\}$,

$\text{∙}$ $f:{𝔽}_{11}^{\text{∗}}\to {𝔽}_{11}^{\text{∗}}$ is injective and ${𝔽}_{11}^{\text{∗}}$ is finite, hence $f$ is bijective.

In ${𝔽}_{11}$, $0=0^3,1=1^3,2=7^3,3=9^3,4=5^3,5=3^3,6=8^3,7=6^3,8=2^3,9=4^3,10=10^3$.

(c)

If $p=13$, then $3\mid p-1,3\wedge (p-1)=3$, so $$\begin{array}{llll}\hfill \exists <!--\; FUNCTION\; APPLICATION\; -->x\in \mathbb{Z},a\equiv x^3(\mathit{mod}13)& ⟺a\equiv 0(\mathit{mod}13)\mathrm{or}{a}^{(p-1)∕3}=a^4\equiv 1(mod13)& \hfill & \\ \hfill & ⟺a\equiv 0,1,-1,5,-5(\mathit{mod}13)& \hfill & \end{array}$$

( $5\equiv 8^3(\mathit{mod}13)$.)