# Prove that if $p \not\equiv 1 \hspace{0.2 cm} (5)$ then $f(x) = x^{5} - 2$ has a unique solution in $\mathbb{F}_{p}$

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.

• Just use that $\phi$ is injective. – Mark Bennet Jan 5 at 20:34
• 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 $p \neq 5$. – Mindlack Jan 5 at 20:39
• So you want to prove that equation $$x^5\equiv _p2$$ has only one solution if $p\ne 1\pmod 5$? – 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.