2
$\begingroup$

To prove the statetament, i thought to define a linear application $$ \phi : \mathbb{F}_{p}^{*} \longmapsto \mathbb{F}_{p}^{*}$$

Define by : $f(x) = x^{5}$, studying the kernel of $\phi$ I noticed that from $x^{5} \equiv 1 \hspace{0.2 cm}(5)$ $\phi$ was injective (Because from the hp. we have $5 \nmid p - 1 $, which translates in $p \not\equiv 1 \hspace{0.2 cm} (5)$).

It follows that this $\phi$ is an automorphism of $\mathbb{F}_{p}^{*}$, in particular it is surjective,

From here i know that for every $p \not\equiv 1 \hspace{0.2 cm} (5)$ it exists $x \in \mathbb{F}_{p}^{*}$ such that $x^{5} = 2$.

Now i know my $f(x) = (x - \alpha)p(x)$ splits into one linear factor, and one of $deg p = 4$.

I'd like to conclude saying that $p$ doesn't split and so that the solution is unique, but don't see how,

Any help or tip would be appreciated,

Thanks.

$\endgroup$
  • 1
    $\begingroup$ Just use that $\phi$ is injective. $\endgroup$ – Mark Bennet Jan 5 at 20:34
  • $\begingroup$ Since $x \in \mathbb{F}_p^* \longmapsto x^5 \in \mathbb{F}_p^*$ is bijective, $f$ is a bijection from $\mathbb{F}_p$ to itself. So it has a unique root. Compare $f$ with its derivative to prove that the root is simple iff $p \neq 5$. $\endgroup$ – Mindlack Jan 5 at 20:39
  • $\begingroup$ So you want to prove that equation $$x^5\equiv _p2$$ has only one solution if $p\ne 1\pmod 5$? $\endgroup$ – Maria Mazur Jan 5 at 20:44
1
$\begingroup$

Hint If $\alpha$ is a multiple root, then $f'(\alpha)=0$.

Alternate solution Use the fact that there exists a primitive root $b$ modulo $p$. Write $x=b^k, 2=b^\alpha$ and solve for $k$.

$\endgroup$
1
$\begingroup$

Let $p= 5k+r$ where $r\in\{0,2,3,4\}$. Let us prove that $$\{0^5-2,1^5-2,...,(p-1)^5-2\}=_{\pmod p} \{0,1,2,...,p-1\}$$

Say there exist $a\ne b \in \mathbb{F}_p$ such that $$a^5-2\equiv _p b^5-2 \implies a^5\equiv _pb^5$$ Since by Fermat theorem we have $a^{5k+r-1}\equiv _p1$ we deduce:$$a^{5k}\equiv _pb^{5k}\implies a^{r-1}\equiv_pb^{r-1} $$

Case 1: $r=0$ (so $p=5$) then $a^5\equiv_5 a$ and $b^5\equiv_5 b$ so $a\equiv _5b$ a contradiciton since $a\ne b$.

Case 2: $r=2$ then $a\equiv_p b$ a contradiciton since $a\ne b$.

Case 3: $r=3$ then $a^2\equiv_p b^2$, then since $a\ne b$ we have $a\equiv_p -b$ but then $a^5\equiv_5 -b^5 \implies p\mid 2a^5 \implies p\mid a \implies p\mid b \implies a=b$ a contradiction.

Case 4: $r=4$ then $a^3\equiv_p b^3$, then $a^6\equiv_5 b^6 \equiv a^5b \implies p\mid a^5(a-b) \implies p\mid a \implies p\mid b \implies a=b$ a contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.