Prove that if $p notequiv 1 hspace{0.2 cm} (5)$ then $f(x) = x^{5} - 2$ has a unique solution in...












2














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 $p notequiv 1 hspace{0.2 cm} (5)$).



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



From here i know that for every $p notequiv 1 hspace{0.2 cm} (5)$ 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.






group-theory field-theory finite-fields splitting-field automorphism-group







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  • 1




    Just use that $phi$ is injective.
    – Mark Bennet
    2 days ago










  • 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
    2 days ago










  • So you want to prove that equation $$x^5equiv _p2$$ has only one solution if $pne 1pmod 5$?
    – greedoid
    2 days ago
















2














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 $p notequiv 1 hspace{0.2 cm} (5)$).



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



From here i know that for every $p notequiv 1 hspace{0.2 cm} (5)$ 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.






group-theory field-theory finite-fields splitting-field automorphism-group







share|cite|improve this question


















  • 1




    Just use that $phi$ is injective.
    – Mark Bennet
    2 days ago










  • 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
    2 days ago










  • So you want to prove that equation $$x^5equiv _p2$$ has only one solution if $pne 1pmod 5$?
    – greedoid
    2 days ago














2












2








2







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 $p notequiv 1 hspace{0.2 cm} (5)$).



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



From here i know that for every $p notequiv 1 hspace{0.2 cm} (5)$ 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.






group-theory field-theory finite-fields splitting-field automorphism-group







share|cite|improve this question













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 $p notequiv 1 hspace{0.2 cm} (5)$).



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



From here i know that for every $p notequiv 1 hspace{0.2 cm} (5)$ 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.






group-theory field-theory finite-fields splitting-field automorphism-group




group-theory field-theory finite-fields splitting-field automorphism-group






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asked 2 days ago









jacopoburellijacopoburelli

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  • 1




    Just use that $phi$ is injective.
    – Mark Bennet
    2 days ago










  • 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
    2 days ago










  • So you want to prove that equation $$x^5equiv _p2$$ has only one solution if $pne 1pmod 5$?
    – greedoid
    2 days ago














  • 1




    Just use that $phi$ is injective.
    – Mark Bennet
    2 days ago










  • 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
    2 days ago










  • So you want to prove that equation $$x^5equiv _p2$$ has only one solution if $pne 1pmod 5$?
    – greedoid
    2 days ago








1




1




Just use that $phi$ is injective.
– Mark Bennet
2 days ago




Just use that $phi$ is injective.
– Mark Bennet
2 days ago












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
2 days ago




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
2 days ago












So you want to prove that equation $$x^5equiv _p2$$ has only one solution if $pne 1pmod 5$?
– greedoid
2 days ago




So you want to prove that equation $$x^5equiv _p2$$ has only one solution if $pne 1pmod 5$?
– greedoid
2 days ago










2 Answers
2






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oldest

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1














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 $k$.






share|cite|improve this answer





























    1














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



    Say there exist $ane b in mathbb{F}_p$ such that $$a^5-2equiv _p b^5-2 implies a^5equiv _pb^5$$
    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 $a^5equiv_5 a$ and $b^5equiv_5 b$ so $aequiv _5b$ a contradiciton since $ane b$.



    Case 2: $r=2$ then $aequiv_p b$ a contradiciton since $ane b$.



    Case 3: $r=3$ then $a^2equiv_p b^2$, then since $ane b$ we have $aequiv_p -b$ but then $a^5equiv_5 -b^5 implies pmid 2a^5 implies pmid a implies pmid b implies a=b$ a contradiction.



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






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

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      1














      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 $k$.






      share|cite|improve this answer


























        1














        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 $k$.






        share|cite|improve this answer
























          1












          1








          1






          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 $k$.






          share|cite|improve this answer












          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 $k$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          N. S.N. S.

          102k5111207




          102k5111207























              1














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



              Say there exist $ane b in mathbb{F}_p$ such that $$a^5-2equiv _p b^5-2 implies a^5equiv _pb^5$$
              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 $a^5equiv_5 a$ and $b^5equiv_5 b$ so $aequiv _5b$ a contradiciton since $ane b$.



              Case 2: $r=2$ then $aequiv_p b$ a contradiciton since $ane b$.



              Case 3: $r=3$ then $a^2equiv_p b^2$, then since $ane b$ we have $aequiv_p -b$ but then $a^5equiv_5 -b^5 implies pmid 2a^5 implies pmid a implies pmid b implies a=b$ a contradiction.



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






              share|cite|improve this answer


























                1














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



                Say there exist $ane b in mathbb{F}_p$ such that $$a^5-2equiv _p b^5-2 implies a^5equiv _pb^5$$
                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 $a^5equiv_5 a$ and $b^5equiv_5 b$ so $aequiv _5b$ a contradiciton since $ane b$.



                Case 2: $r=2$ then $aequiv_p b$ a contradiciton since $ane b$.



                Case 3: $r=3$ then $a^2equiv_p b^2$, then since $ane b$ we have $aequiv_p -b$ but then $a^5equiv_5 -b^5 implies pmid 2a^5 implies pmid a implies pmid b implies a=b$ a contradiction.



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






                share|cite|improve this answer
























                  1












                  1








                  1






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



                  Say there exist $ane b in mathbb{F}_p$ such that $$a^5-2equiv _p b^5-2 implies a^5equiv _pb^5$$
                  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 $a^5equiv_5 a$ and $b^5equiv_5 b$ so $aequiv _5b$ a contradiciton since $ane b$.



                  Case 2: $r=2$ then $aequiv_p b$ a contradiciton since $ane b$.



                  Case 3: $r=3$ then $a^2equiv_p b^2$, then since $ane b$ we have $aequiv_p -b$ but then $a^5equiv_5 -b^5 implies pmid 2a^5 implies pmid a implies pmid b implies a=b$ a contradiction.



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






                  share|cite|improve this answer












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



                  Say there exist $ane b in mathbb{F}_p$ such that $$a^5-2equiv _p b^5-2 implies a^5equiv _pb^5$$
                  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 $a^5equiv_5 a$ and $b^5equiv_5 b$ so $aequiv _5b$ a contradiciton since $ane b$.



                  Case 2: $r=2$ then $aequiv_p b$ a contradiciton since $ane b$.



                  Case 3: $r=3$ then $a^2equiv_p b^2$, then since $ane b$ we have $aequiv_p -b$ but then $a^5equiv_5 -b^5 implies pmid 2a^5 implies pmid a implies pmid b implies a=b$ a contradiction.



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







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 days ago









                  greedoidgreedoid

                  38.4k114797




                  38.4k114797






























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