True or false, there exist exactly two elements $[x]$ in $mathbb{Z}_{11}$ such that $[x]^2 = [2]$. [closed]












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True or false, there exist exactly two elements $[x]$ in $mathbb{Z}_{11}$ such that $[x]^2 = [2]$.



Do I use Fermat's Little Theorem or its corollary?










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closed as off-topic by user21820, user10354138, José Carlos Santos, Gibbs, John B Dec 10 '18 at 11:40


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, user10354138, José Carlos Santos, Gibbs, John B

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    Only 11? Just check them all
    $endgroup$
    – Randall
    Dec 9 '18 at 4:42






  • 1




    $begingroup$
    Have you learnt about the Legendre symbol yet?
    $endgroup$
    – Lord Shark the Unknown
    Dec 9 '18 at 4:43






  • 1




    $begingroup$
    You only need to check half since $x^2 = (-x)^2$
    $endgroup$
    – qwr
    Dec 9 '18 at 4:55
















0












$begingroup$


True or false, there exist exactly two elements $[x]$ in $mathbb{Z}_{11}$ such that $[x]^2 = [2]$.



Do I use Fermat's Little Theorem or its corollary?










share|cite|improve this question









$endgroup$



closed as off-topic by user21820, user10354138, José Carlos Santos, Gibbs, John B Dec 10 '18 at 11:40


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, user10354138, José Carlos Santos, Gibbs, John B

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    Only 11? Just check them all
    $endgroup$
    – Randall
    Dec 9 '18 at 4:42






  • 1




    $begingroup$
    Have you learnt about the Legendre symbol yet?
    $endgroup$
    – Lord Shark the Unknown
    Dec 9 '18 at 4:43






  • 1




    $begingroup$
    You only need to check half since $x^2 = (-x)^2$
    $endgroup$
    – qwr
    Dec 9 '18 at 4:55














0












0








0





$begingroup$


True or false, there exist exactly two elements $[x]$ in $mathbb{Z}_{11}$ such that $[x]^2 = [2]$.



Do I use Fermat's Little Theorem or its corollary?










share|cite|improve this question









$endgroup$




True or false, there exist exactly two elements $[x]$ in $mathbb{Z}_{11}$ such that $[x]^2 = [2]$.



Do I use Fermat's Little Theorem or its corollary?







modular-arithmetic






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 9 '18 at 4:38









hunnybunshunnybuns

15




15




closed as off-topic by user21820, user10354138, José Carlos Santos, Gibbs, John B Dec 10 '18 at 11:40


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, user10354138, José Carlos Santos, Gibbs, John B

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by user21820, user10354138, José Carlos Santos, Gibbs, John B Dec 10 '18 at 11:40


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, user10354138, José Carlos Santos, Gibbs, John B

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    $begingroup$
    Only 11? Just check them all
    $endgroup$
    – Randall
    Dec 9 '18 at 4:42






  • 1




    $begingroup$
    Have you learnt about the Legendre symbol yet?
    $endgroup$
    – Lord Shark the Unknown
    Dec 9 '18 at 4:43






  • 1




    $begingroup$
    You only need to check half since $x^2 = (-x)^2$
    $endgroup$
    – qwr
    Dec 9 '18 at 4:55














  • 2




    $begingroup$
    Only 11? Just check them all
    $endgroup$
    – Randall
    Dec 9 '18 at 4:42






  • 1




    $begingroup$
    Have you learnt about the Legendre symbol yet?
    $endgroup$
    – Lord Shark the Unknown
    Dec 9 '18 at 4:43






  • 1




    $begingroup$
    You only need to check half since $x^2 = (-x)^2$
    $endgroup$
    – qwr
    Dec 9 '18 at 4:55








2




2




$begingroup$
Only 11? Just check them all
$endgroup$
– Randall
Dec 9 '18 at 4:42




$begingroup$
Only 11? Just check them all
$endgroup$
– Randall
Dec 9 '18 at 4:42




1




1




$begingroup$
Have you learnt about the Legendre symbol yet?
$endgroup$
– Lord Shark the Unknown
Dec 9 '18 at 4:43




$begingroup$
Have you learnt about the Legendre symbol yet?
$endgroup$
– Lord Shark the Unknown
Dec 9 '18 at 4:43




1




1




$begingroup$
You only need to check half since $x^2 = (-x)^2$
$endgroup$
– qwr
Dec 9 '18 at 4:55




$begingroup$
You only need to check half since $x^2 = (-x)^2$
$endgroup$
– qwr
Dec 9 '18 at 4:55










2 Answers
2






active

oldest

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Checking one at a time is one approach. It's false.



Now using Fermat's little theorem, if $x^2cong2pmod{11}$, it follows that $(x^2)^5cong32pmod{11}cong10pmod{11}$. So again we get that it's false: actually there are no such elements.






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$endgroup$





















    0












    $begingroup$

    Another approach: Observe that every non-zero element of $Bbb{Z}_{11}$ can be written as a power of $2$ (same as saying $2$ is a primitive root or a generator). If such an $x$ exists, then $x=2^k$ for some $k in {1,2, ldots ,10 }$. This implies:
    begin{align*}
    x^2 & equiv 2 pmod{11}\
    2^{2k} & equiv 2 pmod{11}\
    2^{2k-1} & equiv 1 pmod{11} & (because gcd(2,11)=1)
    end{align*}

    But then $10$ must divide $2k-1$, which is impossible. So no such $x$ exists.






    share|cite|improve this answer









    $endgroup$




















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Checking one at a time is one approach. It's false.



      Now using Fermat's little theorem, if $x^2cong2pmod{11}$, it follows that $(x^2)^5cong32pmod{11}cong10pmod{11}$. So again we get that it's false: actually there are no such elements.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        Checking one at a time is one approach. It's false.



        Now using Fermat's little theorem, if $x^2cong2pmod{11}$, it follows that $(x^2)^5cong32pmod{11}cong10pmod{11}$. So again we get that it's false: actually there are no such elements.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          Checking one at a time is one approach. It's false.



          Now using Fermat's little theorem, if $x^2cong2pmod{11}$, it follows that $(x^2)^5cong32pmod{11}cong10pmod{11}$. So again we get that it's false: actually there are no such elements.






          share|cite|improve this answer











          $endgroup$



          Checking one at a time is one approach. It's false.



          Now using Fermat's little theorem, if $x^2cong2pmod{11}$, it follows that $(x^2)^5cong32pmod{11}cong10pmod{11}$. So again we get that it's false: actually there are no such elements.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 9 '18 at 4:55

























          answered Dec 9 '18 at 4:47









          Chris CusterChris Custer

          12.1k3825




          12.1k3825























              0












              $begingroup$

              Another approach: Observe that every non-zero element of $Bbb{Z}_{11}$ can be written as a power of $2$ (same as saying $2$ is a primitive root or a generator). If such an $x$ exists, then $x=2^k$ for some $k in {1,2, ldots ,10 }$. This implies:
              begin{align*}
              x^2 & equiv 2 pmod{11}\
              2^{2k} & equiv 2 pmod{11}\
              2^{2k-1} & equiv 1 pmod{11} & (because gcd(2,11)=1)
              end{align*}

              But then $10$ must divide $2k-1$, which is impossible. So no such $x$ exists.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Another approach: Observe that every non-zero element of $Bbb{Z}_{11}$ can be written as a power of $2$ (same as saying $2$ is a primitive root or a generator). If such an $x$ exists, then $x=2^k$ for some $k in {1,2, ldots ,10 }$. This implies:
                begin{align*}
                x^2 & equiv 2 pmod{11}\
                2^{2k} & equiv 2 pmod{11}\
                2^{2k-1} & equiv 1 pmod{11} & (because gcd(2,11)=1)
                end{align*}

                But then $10$ must divide $2k-1$, which is impossible. So no such $x$ exists.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Another approach: Observe that every non-zero element of $Bbb{Z}_{11}$ can be written as a power of $2$ (same as saying $2$ is a primitive root or a generator). If such an $x$ exists, then $x=2^k$ for some $k in {1,2, ldots ,10 }$. This implies:
                  begin{align*}
                  x^2 & equiv 2 pmod{11}\
                  2^{2k} & equiv 2 pmod{11}\
                  2^{2k-1} & equiv 1 pmod{11} & (because gcd(2,11)=1)
                  end{align*}

                  But then $10$ must divide $2k-1$, which is impossible. So no such $x$ exists.






                  share|cite|improve this answer









                  $endgroup$



                  Another approach: Observe that every non-zero element of $Bbb{Z}_{11}$ can be written as a power of $2$ (same as saying $2$ is a primitive root or a generator). If such an $x$ exists, then $x=2^k$ for some $k in {1,2, ldots ,10 }$. This implies:
                  begin{align*}
                  x^2 & equiv 2 pmod{11}\
                  2^{2k} & equiv 2 pmod{11}\
                  2^{2k-1} & equiv 1 pmod{11} & (because gcd(2,11)=1)
                  end{align*}

                  But then $10$ must divide $2k-1$, which is impossible. So no such $x$ exists.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 9 '18 at 6:00









                  Anurag AAnurag A

                  26k12249




                  26k12249















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