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,


  • 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

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 $k$.


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.


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