Show that if $a$ has order $3bmod p$ then $a+1$ has order $6bmod p$.












2












$begingroup$


Show that if $a$ has order $3bmod p$ then $a+1$ has order $6bmod p$. I know I am supposed to use primitive roots but I think this is where I am getting caught up. The definition of primitive root is "if $a$ is a least residue and the order of $abmod p$ is $phi(m)$ then $a$ is a primitive root of $m$. But I really am not sure how to use this to my advantage in solving anything.



Thanks!






number-theory







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  • $begingroup$
    math.stackexchange.com/questions/332325/…
    $endgroup$
    – Zach Teitler
    Dec 14 '18 at 20:29
















2












$begingroup$


Show that if $a$ has order $3bmod p$ then $a+1$ has order $6bmod p$. I know I am supposed to use primitive roots but I think this is where I am getting caught up. The definition of primitive root is "if $a$ is a least residue and the order of $abmod p$ is $phi(m)$ then $a$ is a primitive root of $m$. But I really am not sure how to use this to my advantage in solving anything.



Thanks!






number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    math.stackexchange.com/questions/332325/…
    $endgroup$
    – Zach Teitler
    Dec 14 '18 at 20:29














2












2








2





$begingroup$


Show that if $a$ has order $3bmod p$ then $a+1$ has order $6bmod p$. I know I am supposed to use primitive roots but I think this is where I am getting caught up. The definition of primitive root is "if $a$ is a least residue and the order of $abmod p$ is $phi(m)$ then $a$ is a primitive root of $m$. But I really am not sure how to use this to my advantage in solving anything.



Thanks!






number-theory







share|cite|improve this question











$endgroup$




Show that if $a$ has order $3bmod p$ then $a+1$ has order $6bmod p$. I know I am supposed to use primitive roots but I think this is where I am getting caught up. The definition of primitive root is "if $a$ is a least residue and the order of $abmod p$ is $phi(m)$ then $a$ is a primitive root of $m$. But I really am not sure how to use this to my advantage in solving anything.



Thanks!






number-theory




number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 8 '13 at 19:05









MvG

31k450105




31k450105










asked Aug 8 '13 at 18:08









KathyKathy

1019




1019












  • $begingroup$
    math.stackexchange.com/questions/332325/…
    $endgroup$
    – Zach Teitler
    Dec 14 '18 at 20:29


















  • $begingroup$
    math.stackexchange.com/questions/332325/…
    $endgroup$
    – Zach Teitler
    Dec 14 '18 at 20:29
















$begingroup$
math.stackexchange.com/questions/332325/…
$endgroup$
– Zach Teitler
Dec 14 '18 at 20:29




$begingroup$
math.stackexchange.com/questions/332325/…
$endgroup$
– Zach Teitler
Dec 14 '18 at 20:29










2 Answers
2






active

oldest

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9












$begingroup$

Note that if $a=1$, $a$ has order $1$. Thus, we can assume $ane1$. Furthermore, $pne2$ since no element mod $2$ has order $3$. Therefore, $-1ne1pmod{p}$.
$$
begin{align}
(a+1)^3
&=a^3+3a^2+3a+1\
&=1+3a^2+3a+1\
&=-1+3(1+a+a^2)\
&=-1+3frac{a^3-1}{a-1}\
&=-1
end{align}
$$
Therefore, $(a+1)^3=-1$ and $(a+1)^6=1$. (Shortened a la ccorn).
$$
begin{align}
(a+1)^2
&=a^2+2a+1\
&=a+(a^2+a+1)\
&=a+frac{a^3-1}{a-1}\
&=a\
&ne1
end{align}
$$
Therefore, $(a+1)^2ne1$.



Thus, $(a+1)$ has order $6$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    $(a+1)^6=1$ also follows easily from $(a+1)^3=-1$
    $endgroup$
    – ccorn
    Aug 8 '13 at 18:25






  • 1




    $begingroup$
    @ccorn: indeed it does, but I had already written the part for $(a+1)^6$. I guess I could shorten the answer. Thanks.
    $endgroup$
    – robjohn♦
    Aug 8 '13 at 18:27










  • $begingroup$
    With the edit made the (a+1)^{2} is not needed right? If it is I don't understand.
    $endgroup$
    – Kathy
    Aug 8 '13 at 19:19






  • 1




    $begingroup$
    @Kathy: we still have to show that $a+1$ doesn't have order $2$. We could argue that if $(a+1)^2=1$ and $(a+1)^3=-1$, then $a+1=-1$; that is, $a=-2$. Since $a^3=1$ we have that $-8equiv1pmod{p}$, which implies that $pmid9$, whence $p=3$. Since there are no elements of order $3$ mod $3$, we could stop. Personally, I think the argument in my answer is simpler.
    $endgroup$
    – robjohn♦
    Aug 8 '13 at 19:38










  • $begingroup$
    Is there a problem with this answer? Just wondering if the downvote means something needs fixing.
    $endgroup$
    – robjohn♦
    Jan 1 at 19:45



















1












$begingroup$

ord_$pa=3iff p$ divides $(a^3-1)=(a-1)(a^2+a+1)$



But $pnotmid (a-1)$ as ord_$pa=3 $



$displaystyleimplies a^2+a+1equiv0pmod p$
$displaystyleiff a(a+1)equiv-1pmod p$



Method $1:$



$implies a^3(a+1)^3equiv-1pmod pimplies (a+1)^3equiv-1$ and $(a+1)^6equiv1$



$implies $ord_$p(a+1)mid 6$ but does not divide $3$



If ord_$p(a+1)mid 2, (a+1)^2equiv1pmod p$ and $a^2(a+1)^2equiv1pmod pimplies a^2equiv1$ which is impossible as ord$_pa=3$



$implies $ord$_p(a+1)=6$



Method $2:$



$displaystyle a(a+1)equiv-1pmod piff a+1equiv (-a)^{-1}pmod p$



Again as $a^3equiv1pmod p, (-a)^3=-a^3equiv-1pmod pimplies (-a)^6equiv1implies $ord$_p(-a)=6$



Using this, ord$_p{(-a)^{-1}})=$ord$_p(-a)=6$






share|cite|improve this answer











$endgroup$













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    2 Answers
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    2 Answers
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    active

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    9












    $begingroup$

    Note that if $a=1$, $a$ has order $1$. Thus, we can assume $ane1$. Furthermore, $pne2$ since no element mod $2$ has order $3$. Therefore, $-1ne1pmod{p}$.
    $$
    begin{align}
    (a+1)^3
    &=a^3+3a^2+3a+1\
    &=1+3a^2+3a+1\
    &=-1+3(1+a+a^2)\
    &=-1+3frac{a^3-1}{a-1}\
    &=-1
    end{align}
    $$
    Therefore, $(a+1)^3=-1$ and $(a+1)^6=1$. (Shortened a la ccorn).
    $$
    begin{align}
    (a+1)^2
    &=a^2+2a+1\
    &=a+(a^2+a+1)\
    &=a+frac{a^3-1}{a-1}\
    &=a\
    &ne1
    end{align}
    $$
    Therefore, $(a+1)^2ne1$.



    Thus, $(a+1)$ has order $6$.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      $(a+1)^6=1$ also follows easily from $(a+1)^3=-1$
      $endgroup$
      – ccorn
      Aug 8 '13 at 18:25






    • 1




      $begingroup$
      @ccorn: indeed it does, but I had already written the part for $(a+1)^6$. I guess I could shorten the answer. Thanks.
      $endgroup$
      – robjohn♦
      Aug 8 '13 at 18:27










    • $begingroup$
      With the edit made the (a+1)^{2} is not needed right? If it is I don't understand.
      $endgroup$
      – Kathy
      Aug 8 '13 at 19:19






    • 1




      $begingroup$
      @Kathy: we still have to show that $a+1$ doesn't have order $2$. We could argue that if $(a+1)^2=1$ and $(a+1)^3=-1$, then $a+1=-1$; that is, $a=-2$. Since $a^3=1$ we have that $-8equiv1pmod{p}$, which implies that $pmid9$, whence $p=3$. Since there are no elements of order $3$ mod $3$, we could stop. Personally, I think the argument in my answer is simpler.
      $endgroup$
      – robjohn♦
      Aug 8 '13 at 19:38










    • $begingroup$
      Is there a problem with this answer? Just wondering if the downvote means something needs fixing.
      $endgroup$
      – robjohn♦
      Jan 1 at 19:45
















    9












    $begingroup$

    Note that if $a=1$, $a$ has order $1$. Thus, we can assume $ane1$. Furthermore, $pne2$ since no element mod $2$ has order $3$. Therefore, $-1ne1pmod{p}$.
    $$
    begin{align}
    (a+1)^3
    &=a^3+3a^2+3a+1\
    &=1+3a^2+3a+1\
    &=-1+3(1+a+a^2)\
    &=-1+3frac{a^3-1}{a-1}\
    &=-1
    end{align}
    $$
    Therefore, $(a+1)^3=-1$ and $(a+1)^6=1$. (Shortened a la ccorn).
    $$
    begin{align}
    (a+1)^2
    &=a^2+2a+1\
    &=a+(a^2+a+1)\
    &=a+frac{a^3-1}{a-1}\
    &=a\
    &ne1
    end{align}
    $$
    Therefore, $(a+1)^2ne1$.



    Thus, $(a+1)$ has order $6$.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      $(a+1)^6=1$ also follows easily from $(a+1)^3=-1$
      $endgroup$
      – ccorn
      Aug 8 '13 at 18:25






    • 1




      $begingroup$
      @ccorn: indeed it does, but I had already written the part for $(a+1)^6$. I guess I could shorten the answer. Thanks.
      $endgroup$
      – robjohn♦
      Aug 8 '13 at 18:27










    • $begingroup$
      With the edit made the (a+1)^{2} is not needed right? If it is I don't understand.
      $endgroup$
      – Kathy
      Aug 8 '13 at 19:19






    • 1




      $begingroup$
      @Kathy: we still have to show that $a+1$ doesn't have order $2$. We could argue that if $(a+1)^2=1$ and $(a+1)^3=-1$, then $a+1=-1$; that is, $a=-2$. Since $a^3=1$ we have that $-8equiv1pmod{p}$, which implies that $pmid9$, whence $p=3$. Since there are no elements of order $3$ mod $3$, we could stop. Personally, I think the argument in my answer is simpler.
      $endgroup$
      – robjohn♦
      Aug 8 '13 at 19:38










    • $begingroup$
      Is there a problem with this answer? Just wondering if the downvote means something needs fixing.
      $endgroup$
      – robjohn♦
      Jan 1 at 19:45














    9












    9








    9





    $begingroup$

    Note that if $a=1$, $a$ has order $1$. Thus, we can assume $ane1$. Furthermore, $pne2$ since no element mod $2$ has order $3$. Therefore, $-1ne1pmod{p}$.
    $$
    begin{align}
    (a+1)^3
    &=a^3+3a^2+3a+1\
    &=1+3a^2+3a+1\
    &=-1+3(1+a+a^2)\
    &=-1+3frac{a^3-1}{a-1}\
    &=-1
    end{align}
    $$
    Therefore, $(a+1)^3=-1$ and $(a+1)^6=1$. (Shortened a la ccorn).
    $$
    begin{align}
    (a+1)^2
    &=a^2+2a+1\
    &=a+(a^2+a+1)\
    &=a+frac{a^3-1}{a-1}\
    &=a\
    &ne1
    end{align}
    $$
    Therefore, $(a+1)^2ne1$.



    Thus, $(a+1)$ has order $6$.






    share|cite|improve this answer











    $endgroup$



    Note that if $a=1$, $a$ has order $1$. Thus, we can assume $ane1$. Furthermore, $pne2$ since no element mod $2$ has order $3$. Therefore, $-1ne1pmod{p}$.
    $$
    begin{align}
    (a+1)^3
    &=a^3+3a^2+3a+1\
    &=1+3a^2+3a+1\
    &=-1+3(1+a+a^2)\
    &=-1+3frac{a^3-1}{a-1}\
    &=-1
    end{align}
    $$
    Therefore, $(a+1)^3=-1$ and $(a+1)^6=1$. (Shortened a la ccorn).
    $$
    begin{align}
    (a+1)^2
    &=a^2+2a+1\
    &=a+(a^2+a+1)\
    &=a+frac{a^3-1}{a-1}\
    &=a\
    &ne1
    end{align}
    $$
    Therefore, $(a+1)^2ne1$.



    Thus, $(a+1)$ has order $6$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 1 at 20:27

























    answered Aug 8 '13 at 18:24









    robjohn♦robjohn

    268k27308634




    268k27308634








    • 1




      $begingroup$
      $(a+1)^6=1$ also follows easily from $(a+1)^3=-1$
      $endgroup$
      – ccorn
      Aug 8 '13 at 18:25






    • 1




      $begingroup$
      @ccorn: indeed it does, but I had already written the part for $(a+1)^6$. I guess I could shorten the answer. Thanks.
      $endgroup$
      – robjohn♦
      Aug 8 '13 at 18:27










    • $begingroup$
      With the edit made the (a+1)^{2} is not needed right? If it is I don't understand.
      $endgroup$
      – Kathy
      Aug 8 '13 at 19:19






    • 1




      $begingroup$
      @Kathy: we still have to show that $a+1$ doesn't have order $2$. We could argue that if $(a+1)^2=1$ and $(a+1)^3=-1$, then $a+1=-1$; that is, $a=-2$. Since $a^3=1$ we have that $-8equiv1pmod{p}$, which implies that $pmid9$, whence $p=3$. Since there are no elements of order $3$ mod $3$, we could stop. Personally, I think the argument in my answer is simpler.
      $endgroup$
      – robjohn♦
      Aug 8 '13 at 19:38










    • $begingroup$
      Is there a problem with this answer? Just wondering if the downvote means something needs fixing.
      $endgroup$
      – robjohn♦
      Jan 1 at 19:45














    • 1




      $begingroup$
      $(a+1)^6=1$ also follows easily from $(a+1)^3=-1$
      $endgroup$
      – ccorn
      Aug 8 '13 at 18:25






    • 1




      $begingroup$
      @ccorn: indeed it does, but I had already written the part for $(a+1)^6$. I guess I could shorten the answer. Thanks.
      $endgroup$
      – robjohn♦
      Aug 8 '13 at 18:27










    • $begingroup$
      With the edit made the (a+1)^{2} is not needed right? If it is I don't understand.
      $endgroup$
      – Kathy
      Aug 8 '13 at 19:19






    • 1




      $begingroup$
      @Kathy: we still have to show that $a+1$ doesn't have order $2$. We could argue that if $(a+1)^2=1$ and $(a+1)^3=-1$, then $a+1=-1$; that is, $a=-2$. Since $a^3=1$ we have that $-8equiv1pmod{p}$, which implies that $pmid9$, whence $p=3$. Since there are no elements of order $3$ mod $3$, we could stop. Personally, I think the argument in my answer is simpler.
      $endgroup$
      – robjohn♦
      Aug 8 '13 at 19:38










    • $begingroup$
      Is there a problem with this answer? Just wondering if the downvote means something needs fixing.
      $endgroup$
      – robjohn♦
      Jan 1 at 19:45








    1




    1




    $begingroup$
    $(a+1)^6=1$ also follows easily from $(a+1)^3=-1$
    $endgroup$
    – ccorn
    Aug 8 '13 at 18:25




    $begingroup$
    $(a+1)^6=1$ also follows easily from $(a+1)^3=-1$
    $endgroup$
    – ccorn
    Aug 8 '13 at 18:25




    1




    1




    $begingroup$
    @ccorn: indeed it does, but I had already written the part for $(a+1)^6$. I guess I could shorten the answer. Thanks.
    $endgroup$
    – robjohn♦
    Aug 8 '13 at 18:27




    $begingroup$
    @ccorn: indeed it does, but I had already written the part for $(a+1)^6$. I guess I could shorten the answer. Thanks.
    $endgroup$
    – robjohn♦
    Aug 8 '13 at 18:27












    $begingroup$
    With the edit made the (a+1)^{2} is not needed right? If it is I don't understand.
    $endgroup$
    – Kathy
    Aug 8 '13 at 19:19




    $begingroup$
    With the edit made the (a+1)^{2} is not needed right? If it is I don't understand.
    $endgroup$
    – Kathy
    Aug 8 '13 at 19:19




    1




    1




    $begingroup$
    @Kathy: we still have to show that $a+1$ doesn't have order $2$. We could argue that if $(a+1)^2=1$ and $(a+1)^3=-1$, then $a+1=-1$; that is, $a=-2$. Since $a^3=1$ we have that $-8equiv1pmod{p}$, which implies that $pmid9$, whence $p=3$. Since there are no elements of order $3$ mod $3$, we could stop. Personally, I think the argument in my answer is simpler.
    $endgroup$
    – robjohn♦
    Aug 8 '13 at 19:38




    $begingroup$
    @Kathy: we still have to show that $a+1$ doesn't have order $2$. We could argue that if $(a+1)^2=1$ and $(a+1)^3=-1$, then $a+1=-1$; that is, $a=-2$. Since $a^3=1$ we have that $-8equiv1pmod{p}$, which implies that $pmid9$, whence $p=3$. Since there are no elements of order $3$ mod $3$, we could stop. Personally, I think the argument in my answer is simpler.
    $endgroup$
    – robjohn♦
    Aug 8 '13 at 19:38












    $begingroup$
    Is there a problem with this answer? Just wondering if the downvote means something needs fixing.
    $endgroup$
    – robjohn♦
    Jan 1 at 19:45




    $begingroup$
    Is there a problem with this answer? Just wondering if the downvote means something needs fixing.
    $endgroup$
    – robjohn♦
    Jan 1 at 19:45











    1












    $begingroup$

    ord_$pa=3iff p$ divides $(a^3-1)=(a-1)(a^2+a+1)$



    But $pnotmid (a-1)$ as ord_$pa=3 $



    $displaystyleimplies a^2+a+1equiv0pmod p$
    $displaystyleiff a(a+1)equiv-1pmod p$



    Method $1:$



    $implies a^3(a+1)^3equiv-1pmod pimplies (a+1)^3equiv-1$ and $(a+1)^6equiv1$



    $implies $ord_$p(a+1)mid 6$ but does not divide $3$



    If ord_$p(a+1)mid 2, (a+1)^2equiv1pmod p$ and $a^2(a+1)^2equiv1pmod pimplies a^2equiv1$ which is impossible as ord$_pa=3$



    $implies $ord$_p(a+1)=6$



    Method $2:$



    $displaystyle a(a+1)equiv-1pmod piff a+1equiv (-a)^{-1}pmod p$



    Again as $a^3equiv1pmod p, (-a)^3=-a^3equiv-1pmod pimplies (-a)^6equiv1implies $ord$_p(-a)=6$



    Using this, ord$_p{(-a)^{-1}})=$ord$_p(-a)=6$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      ord_$pa=3iff p$ divides $(a^3-1)=(a-1)(a^2+a+1)$



      But $pnotmid (a-1)$ as ord_$pa=3 $



      $displaystyleimplies a^2+a+1equiv0pmod p$
      $displaystyleiff a(a+1)equiv-1pmod p$



      Method $1:$



      $implies a^3(a+1)^3equiv-1pmod pimplies (a+1)^3equiv-1$ and $(a+1)^6equiv1$



      $implies $ord_$p(a+1)mid 6$ but does not divide $3$



      If ord_$p(a+1)mid 2, (a+1)^2equiv1pmod p$ and $a^2(a+1)^2equiv1pmod pimplies a^2equiv1$ which is impossible as ord$_pa=3$



      $implies $ord$_p(a+1)=6$



      Method $2:$



      $displaystyle a(a+1)equiv-1pmod piff a+1equiv (-a)^{-1}pmod p$



      Again as $a^3equiv1pmod p, (-a)^3=-a^3equiv-1pmod pimplies (-a)^6equiv1implies $ord$_p(-a)=6$



      Using this, ord$_p{(-a)^{-1}})=$ord$_p(-a)=6$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        ord_$pa=3iff p$ divides $(a^3-1)=(a-1)(a^2+a+1)$



        But $pnotmid (a-1)$ as ord_$pa=3 $



        $displaystyleimplies a^2+a+1equiv0pmod p$
        $displaystyleiff a(a+1)equiv-1pmod p$



        Method $1:$



        $implies a^3(a+1)^3equiv-1pmod pimplies (a+1)^3equiv-1$ and $(a+1)^6equiv1$



        $implies $ord_$p(a+1)mid 6$ but does not divide $3$



        If ord_$p(a+1)mid 2, (a+1)^2equiv1pmod p$ and $a^2(a+1)^2equiv1pmod pimplies a^2equiv1$ which is impossible as ord$_pa=3$



        $implies $ord$_p(a+1)=6$



        Method $2:$



        $displaystyle a(a+1)equiv-1pmod piff a+1equiv (-a)^{-1}pmod p$



        Again as $a^3equiv1pmod p, (-a)^3=-a^3equiv-1pmod pimplies (-a)^6equiv1implies $ord$_p(-a)=6$



        Using this, ord$_p{(-a)^{-1}})=$ord$_p(-a)=6$






        share|cite|improve this answer











        $endgroup$



        ord_$pa=3iff p$ divides $(a^3-1)=(a-1)(a^2+a+1)$



        But $pnotmid (a-1)$ as ord_$pa=3 $



        $displaystyleimplies a^2+a+1equiv0pmod p$
        $displaystyleiff a(a+1)equiv-1pmod p$



        Method $1:$



        $implies a^3(a+1)^3equiv-1pmod pimplies (a+1)^3equiv-1$ and $(a+1)^6equiv1$



        $implies $ord_$p(a+1)mid 6$ but does not divide $3$



        If ord_$p(a+1)mid 2, (a+1)^2equiv1pmod p$ and $a^2(a+1)^2equiv1pmod pimplies a^2equiv1$ which is impossible as ord$_pa=3$



        $implies $ord$_p(a+1)=6$



        Method $2:$



        $displaystyle a(a+1)equiv-1pmod piff a+1equiv (-a)^{-1}pmod p$



        Again as $a^3equiv1pmod p, (-a)^3=-a^3equiv-1pmod pimplies (-a)^6equiv1implies $ord$_p(-a)=6$



        Using this, ord$_p{(-a)^{-1}})=$ord$_p(-a)=6$







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        edited Apr 13 '17 at 12:19









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        answered Aug 8 '13 at 18:12









        lab bhattacharjeelab bhattacharjee

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