If $a$ has order 3 $pmod p$, $p$ prime, then $1+a+a^2 equiv 0 pmod p$ and $1+a$ has order $6$












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There is a little proposition left to the readers in my number theory book, and I'm not sure how to answer this in a proper way:




If $a$ has order $3 pmod p$, $p$ prime, then $1+a+a^2 equiv 0 pmod p$ and $1+a$ has order $6$.




Some hints?






abstract-algebra number-theory modular-arithmetic







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    $begingroup$
    If $a$ has order three modulo $p$, then $a^3equiv1$ but $anotequiv1pmod p$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 28 '18 at 17:02








  • 1




    $begingroup$
    math.stackexchange.com/questions/220493/…
    $endgroup$
    – lab bhattacharjee
    Dec 29 '18 at 2:40
















1












$begingroup$


There is a little proposition left to the readers in my number theory book, and I'm not sure how to answer this in a proper way:




If $a$ has order $3 pmod p$, $p$ prime, then $1+a+a^2 equiv 0 pmod p$ and $1+a$ has order $6$.




Some hints?






abstract-algebra number-theory modular-arithmetic







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    If $a$ has order three modulo $p$, then $a^3equiv1$ but $anotequiv1pmod p$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 28 '18 at 17:02








  • 1




    $begingroup$
    math.stackexchange.com/questions/220493/…
    $endgroup$
    – lab bhattacharjee
    Dec 29 '18 at 2:40














1












1








1


1



$begingroup$


There is a little proposition left to the readers in my number theory book, and I'm not sure how to answer this in a proper way:




If $a$ has order $3 pmod p$, $p$ prime, then $1+a+a^2 equiv 0 pmod p$ and $1+a$ has order $6$.




Some hints?






abstract-algebra number-theory modular-arithmetic







share|cite|improve this question











$endgroup$




There is a little proposition left to the readers in my number theory book, and I'm not sure how to answer this in a proper way:




If $a$ has order $3 pmod p$, $p$ prime, then $1+a+a^2 equiv 0 pmod p$ and $1+a$ has order $6$.




Some hints?






abstract-algebra number-theory modular-arithmetic




abstract-algebra number-theory modular-arithmetic






share|cite|improve this question















share|cite|improve this question













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edited Dec 28 '18 at 17:12









Namaste

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asked Dec 28 '18 at 17:00









AlessarAlessar

313115




313115








  • 3




    $begingroup$
    If $a$ has order three modulo $p$, then $a^3equiv1$ but $anotequiv1pmod p$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 28 '18 at 17:02








  • 1




    $begingroup$
    math.stackexchange.com/questions/220493/…
    $endgroup$
    – lab bhattacharjee
    Dec 29 '18 at 2:40














  • 3




    $begingroup$
    If $a$ has order three modulo $p$, then $a^3equiv1$ but $anotequiv1pmod p$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 28 '18 at 17:02








  • 1




    $begingroup$
    math.stackexchange.com/questions/220493/…
    $endgroup$
    – lab bhattacharjee
    Dec 29 '18 at 2:40








3




3




$begingroup$
If $a$ has order three modulo $p$, then $a^3equiv1$ but $anotequiv1pmod p$.
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 17:02






$begingroup$
If $a$ has order three modulo $p$, then $a^3equiv1$ but $anotequiv1pmod p$.
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 17:02






1




1




$begingroup$
math.stackexchange.com/questions/220493/…
$endgroup$
– lab bhattacharjee
Dec 29 '18 at 2:40




$begingroup$
math.stackexchange.com/questions/220493/…
$endgroup$
– lab bhattacharjee
Dec 29 '18 at 2:40










4 Answers
4






active

oldest

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1












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How $a$ has order $3$, then $anotequiv 1$.



Now, note that $(a-1)(1+a+a^2) = a^3-1 equiv 0$ (mod $p$). As $mathbb{Z}_p$ is a field and $a-1neq 0$, so $1+a+a^2=0$ (mod $p$).



To see why $1+a$ has order $6$, note that
$$(1+a)^6=(1+2a+a^2)^3=a^3=1$$
So, the order of $1+a$ divide $6$.



If $1+a$ has order $1$, then $1+a=1Rightarrow a=0$. This contradicts the fact that $a$ has order $3$.



If $1+a$ has order $2$, then $(1+a)^2=1Rightarrow 1+2a+a^2=1Rightarrow a=1$. But $1$ has order $1$, not $3$.



If $1+a$ has order $3$, then $(1+a)^3=1Rightarrow (1+a)(1+2a+a^2)=1Rightarrow (1+a)a=1Rightarrow a+a^2=1Rightarrow 1+a+a^2=2Rightarrow 0=2$, so $p=2$. But $mathbb{Z}_2$ has no elements with order $3$.



So, the only possible order for $1+a$ is $6$.






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  • $begingroup$
    About the order 2, can you clarify the passage from this $1+2a+a^2=1Rightarrow a=1$ that leads to the conclusion?
    $endgroup$
    – Alessar
    Dec 29 '18 at 8:10










  • $begingroup$
    Also for the part when you say $(1+2a+a^2)^3=a^3$ maybe! I think that it's because there is some definition for $1+2a+a^2=a$?
    $endgroup$
    – Alessar
    Dec 29 '18 at 8:52






  • 1




    $begingroup$
    @Alessar All use $,(1+a)^2 = a,$ (by adding $,a,$ to $,1+a+a^2 = 0) $ It's slightly easier to use powers as in my answer. The reason I use the Order Test there (vs. checking all divisors) is that this will be much quicker for larger numbers/
    $endgroup$
    – Bill Dubuque
    Dec 29 '18 at 14:17





















5












$begingroup$

Work in the residue class ring $bf Z/pbf Z$, which is a field. As in any commutative ring, $x^3-1=(x-1)(x^2+x+1)$, and $a-1ne 0$ if $a$ has order $3$.



For the second assertion, $1+a=-a^2$. If $a$ has order $3$, can deduce the order of $-a^2$?






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    $!bmod p!:, a^3equiv 1, anotequiv 1,$ so $,0equiv a^3-1equiv (a-1)(a^2+a+1),Rightarrow,a^2+a+1equiv 0$



    Thus $ 1+aequiv -a^2. $ To compute $,(1+a)^n,$ we power this congruence:



    Thus $,(1+a)^6equiv, a^{12}equiv 1, $ by $,a^3equiv 1$



    But $ color{#0a0}{(1+a)^2}equiv a^4, equiv, a, color{#0a0}{notequiv 1},$ by hypothesis



    and $ , color{#0a0}{(1+a)^3}!equiv! -a^6!equiv! -1 color{#0a0}{notequiv 1},$ by $,pneq 2, $ (where only order $1$ is possible)



    Thus $,1+a,$ has order $,6,$ by the Order Test below, with $,color{#0a0}{p = 2,3}$



    Order Test $ ,a,$ has order $,n iff a^{large n} = 1 $ but $ color{#0a0}{a^{large n/p} neq 1},$ for every prime $,pmid n$



    Proof $, (Leftarrow), $ Let $,a,$ have $,color{#c00}{{rm order} k}.,$ Then $,kmid n,$ (proof). $ $ If $:k < n,$ then $,k,$ is proper divisor of $,n,$ so by unique factorization $,k,$ arises by deleting at least one prime $,p,$ from the prime factorization of $,n,,$ so $,kmid n/p,,$ say $, kj = n/p, $ so $ color{#0a0}{a^{large n/p}} = (color{#c00}{a^{large k}})^{large j} = color{#c00}1^{large j} = color{#0a0}1,$ contra $rmcolor{#0a0}{hypothesis}$. So $,k=n.$ $, (Rightarrow) $ Clear






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      3












      $begingroup$

      As $a$ has order $3$, $a^3equiv 1 pmod p$. Hence $a^3-1 equiv 0 pmod p$.
      So $a^3-1=(a-1)(1+a+a^2)equiv 0 pmod p$.



      Now, $(1+a)^2=1+2a+a^2$. This will be a nice hint. Use the previous part in some way.






      share|cite|improve this answer









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        4 Answers
        4






        active

        oldest

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        4 Answers
        4






        active

        oldest

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        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        How $a$ has order $3$, then $anotequiv 1$.



        Now, note that $(a-1)(1+a+a^2) = a^3-1 equiv 0$ (mod $p$). As $mathbb{Z}_p$ is a field and $a-1neq 0$, so $1+a+a^2=0$ (mod $p$).



        To see why $1+a$ has order $6$, note that
        $$(1+a)^6=(1+2a+a^2)^3=a^3=1$$
        So, the order of $1+a$ divide $6$.



        If $1+a$ has order $1$, then $1+a=1Rightarrow a=0$. This contradicts the fact that $a$ has order $3$.



        If $1+a$ has order $2$, then $(1+a)^2=1Rightarrow 1+2a+a^2=1Rightarrow a=1$. But $1$ has order $1$, not $3$.



        If $1+a$ has order $3$, then $(1+a)^3=1Rightarrow (1+a)(1+2a+a^2)=1Rightarrow (1+a)a=1Rightarrow a+a^2=1Rightarrow 1+a+a^2=2Rightarrow 0=2$, so $p=2$. But $mathbb{Z}_2$ has no elements with order $3$.



        So, the only possible order for $1+a$ is $6$.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          About the order 2, can you clarify the passage from this $1+2a+a^2=1Rightarrow a=1$ that leads to the conclusion?
          $endgroup$
          – Alessar
          Dec 29 '18 at 8:10










        • $begingroup$
          Also for the part when you say $(1+2a+a^2)^3=a^3$ maybe! I think that it's because there is some definition for $1+2a+a^2=a$?
          $endgroup$
          – Alessar
          Dec 29 '18 at 8:52






        • 1




          $begingroup$
          @Alessar All use $,(1+a)^2 = a,$ (by adding $,a,$ to $,1+a+a^2 = 0) $ It's slightly easier to use powers as in my answer. The reason I use the Order Test there (vs. checking all divisors) is that this will be much quicker for larger numbers/
          $endgroup$
          – Bill Dubuque
          Dec 29 '18 at 14:17


















        1












        $begingroup$

        How $a$ has order $3$, then $anotequiv 1$.



        Now, note that $(a-1)(1+a+a^2) = a^3-1 equiv 0$ (mod $p$). As $mathbb{Z}_p$ is a field and $a-1neq 0$, so $1+a+a^2=0$ (mod $p$).



        To see why $1+a$ has order $6$, note that
        $$(1+a)^6=(1+2a+a^2)^3=a^3=1$$
        So, the order of $1+a$ divide $6$.



        If $1+a$ has order $1$, then $1+a=1Rightarrow a=0$. This contradicts the fact that $a$ has order $3$.



        If $1+a$ has order $2$, then $(1+a)^2=1Rightarrow 1+2a+a^2=1Rightarrow a=1$. But $1$ has order $1$, not $3$.



        If $1+a$ has order $3$, then $(1+a)^3=1Rightarrow (1+a)(1+2a+a^2)=1Rightarrow (1+a)a=1Rightarrow a+a^2=1Rightarrow 1+a+a^2=2Rightarrow 0=2$, so $p=2$. But $mathbb{Z}_2$ has no elements with order $3$.



        So, the only possible order for $1+a$ is $6$.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          About the order 2, can you clarify the passage from this $1+2a+a^2=1Rightarrow a=1$ that leads to the conclusion?
          $endgroup$
          – Alessar
          Dec 29 '18 at 8:10










        • $begingroup$
          Also for the part when you say $(1+2a+a^2)^3=a^3$ maybe! I think that it's because there is some definition for $1+2a+a^2=a$?
          $endgroup$
          – Alessar
          Dec 29 '18 at 8:52






        • 1




          $begingroup$
          @Alessar All use $,(1+a)^2 = a,$ (by adding $,a,$ to $,1+a+a^2 = 0) $ It's slightly easier to use powers as in my answer. The reason I use the Order Test there (vs. checking all divisors) is that this will be much quicker for larger numbers/
          $endgroup$
          – Bill Dubuque
          Dec 29 '18 at 14:17
















        1












        1








        1





        $begingroup$

        How $a$ has order $3$, then $anotequiv 1$.



        Now, note that $(a-1)(1+a+a^2) = a^3-1 equiv 0$ (mod $p$). As $mathbb{Z}_p$ is a field and $a-1neq 0$, so $1+a+a^2=0$ (mod $p$).



        To see why $1+a$ has order $6$, note that
        $$(1+a)^6=(1+2a+a^2)^3=a^3=1$$
        So, the order of $1+a$ divide $6$.



        If $1+a$ has order $1$, then $1+a=1Rightarrow a=0$. This contradicts the fact that $a$ has order $3$.



        If $1+a$ has order $2$, then $(1+a)^2=1Rightarrow 1+2a+a^2=1Rightarrow a=1$. But $1$ has order $1$, not $3$.



        If $1+a$ has order $3$, then $(1+a)^3=1Rightarrow (1+a)(1+2a+a^2)=1Rightarrow (1+a)a=1Rightarrow a+a^2=1Rightarrow 1+a+a^2=2Rightarrow 0=2$, so $p=2$. But $mathbb{Z}_2$ has no elements with order $3$.



        So, the only possible order for $1+a$ is $6$.






        share|cite|improve this answer









        $endgroup$



        How $a$ has order $3$, then $anotequiv 1$.



        Now, note that $(a-1)(1+a+a^2) = a^3-1 equiv 0$ (mod $p$). As $mathbb{Z}_p$ is a field and $a-1neq 0$, so $1+a+a^2=0$ (mod $p$).



        To see why $1+a$ has order $6$, note that
        $$(1+a)^6=(1+2a+a^2)^3=a^3=1$$
        So, the order of $1+a$ divide $6$.



        If $1+a$ has order $1$, then $1+a=1Rightarrow a=0$. This contradicts the fact that $a$ has order $3$.



        If $1+a$ has order $2$, then $(1+a)^2=1Rightarrow 1+2a+a^2=1Rightarrow a=1$. But $1$ has order $1$, not $3$.



        If $1+a$ has order $3$, then $(1+a)^3=1Rightarrow (1+a)(1+2a+a^2)=1Rightarrow (1+a)a=1Rightarrow a+a^2=1Rightarrow 1+a+a^2=2Rightarrow 0=2$, so $p=2$. But $mathbb{Z}_2$ has no elements with order $3$.



        So, the only possible order for $1+a$ is $6$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 28 '18 at 17:30









        Tiago Emilio SillerTiago Emilio Siller

        7401419




        7401419












        • $begingroup$
          About the order 2, can you clarify the passage from this $1+2a+a^2=1Rightarrow a=1$ that leads to the conclusion?
          $endgroup$
          – Alessar
          Dec 29 '18 at 8:10










        • $begingroup$
          Also for the part when you say $(1+2a+a^2)^3=a^3$ maybe! I think that it's because there is some definition for $1+2a+a^2=a$?
          $endgroup$
          – Alessar
          Dec 29 '18 at 8:52






        • 1




          $begingroup$
          @Alessar All use $,(1+a)^2 = a,$ (by adding $,a,$ to $,1+a+a^2 = 0) $ It's slightly easier to use powers as in my answer. The reason I use the Order Test there (vs. checking all divisors) is that this will be much quicker for larger numbers/
          $endgroup$
          – Bill Dubuque
          Dec 29 '18 at 14:17




















        • $begingroup$
          About the order 2, can you clarify the passage from this $1+2a+a^2=1Rightarrow a=1$ that leads to the conclusion?
          $endgroup$
          – Alessar
          Dec 29 '18 at 8:10










        • $begingroup$
          Also for the part when you say $(1+2a+a^2)^3=a^3$ maybe! I think that it's because there is some definition for $1+2a+a^2=a$?
          $endgroup$
          – Alessar
          Dec 29 '18 at 8:52






        • 1




          $begingroup$
          @Alessar All use $,(1+a)^2 = a,$ (by adding $,a,$ to $,1+a+a^2 = 0) $ It's slightly easier to use powers as in my answer. The reason I use the Order Test there (vs. checking all divisors) is that this will be much quicker for larger numbers/
          $endgroup$
          – Bill Dubuque
          Dec 29 '18 at 14:17


















        $begingroup$
        About the order 2, can you clarify the passage from this $1+2a+a^2=1Rightarrow a=1$ that leads to the conclusion?
        $endgroup$
        – Alessar
        Dec 29 '18 at 8:10




        $begingroup$
        About the order 2, can you clarify the passage from this $1+2a+a^2=1Rightarrow a=1$ that leads to the conclusion?
        $endgroup$
        – Alessar
        Dec 29 '18 at 8:10












        $begingroup$
        Also for the part when you say $(1+2a+a^2)^3=a^3$ maybe! I think that it's because there is some definition for $1+2a+a^2=a$?
        $endgroup$
        – Alessar
        Dec 29 '18 at 8:52




        $begingroup$
        Also for the part when you say $(1+2a+a^2)^3=a^3$ maybe! I think that it's because there is some definition for $1+2a+a^2=a$?
        $endgroup$
        – Alessar
        Dec 29 '18 at 8:52




        1




        1




        $begingroup$
        @Alessar All use $,(1+a)^2 = a,$ (by adding $,a,$ to $,1+a+a^2 = 0) $ It's slightly easier to use powers as in my answer. The reason I use the Order Test there (vs. checking all divisors) is that this will be much quicker for larger numbers/
        $endgroup$
        – Bill Dubuque
        Dec 29 '18 at 14:17






        $begingroup$
        @Alessar All use $,(1+a)^2 = a,$ (by adding $,a,$ to $,1+a+a^2 = 0) $ It's slightly easier to use powers as in my answer. The reason I use the Order Test there (vs. checking all divisors) is that this will be much quicker for larger numbers/
        $endgroup$
        – Bill Dubuque
        Dec 29 '18 at 14:17













        5












        $begingroup$

        Work in the residue class ring $bf Z/pbf Z$, which is a field. As in any commutative ring, $x^3-1=(x-1)(x^2+x+1)$, and $a-1ne 0$ if $a$ has order $3$.



        For the second assertion, $1+a=-a^2$. If $a$ has order $3$, can deduce the order of $-a^2$?






        share|cite|improve this answer









        $endgroup$


















          5












          $begingroup$

          Work in the residue class ring $bf Z/pbf Z$, which is a field. As in any commutative ring, $x^3-1=(x-1)(x^2+x+1)$, and $a-1ne 0$ if $a$ has order $3$.



          For the second assertion, $1+a=-a^2$. If $a$ has order $3$, can deduce the order of $-a^2$?






          share|cite|improve this answer









          $endgroup$
















            5












            5








            5





            $begingroup$

            Work in the residue class ring $bf Z/pbf Z$, which is a field. As in any commutative ring, $x^3-1=(x-1)(x^2+x+1)$, and $a-1ne 0$ if $a$ has order $3$.



            For the second assertion, $1+a=-a^2$. If $a$ has order $3$, can deduce the order of $-a^2$?






            share|cite|improve this answer









            $endgroup$



            Work in the residue class ring $bf Z/pbf Z$, which is a field. As in any commutative ring, $x^3-1=(x-1)(x^2+x+1)$, and $a-1ne 0$ if $a$ has order $3$.



            For the second assertion, $1+a=-a^2$. If $a$ has order $3$, can deduce the order of $-a^2$?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 28 '18 at 17:12









            BernardBernard

            123k741116




            123k741116























                4












                $begingroup$

                $!bmod p!:, a^3equiv 1, anotequiv 1,$ so $,0equiv a^3-1equiv (a-1)(a^2+a+1),Rightarrow,a^2+a+1equiv 0$



                Thus $ 1+aequiv -a^2. $ To compute $,(1+a)^n,$ we power this congruence:



                Thus $,(1+a)^6equiv, a^{12}equiv 1, $ by $,a^3equiv 1$



                But $ color{#0a0}{(1+a)^2}equiv a^4, equiv, a, color{#0a0}{notequiv 1},$ by hypothesis



                and $ , color{#0a0}{(1+a)^3}!equiv! -a^6!equiv! -1 color{#0a0}{notequiv 1},$ by $,pneq 2, $ (where only order $1$ is possible)



                Thus $,1+a,$ has order $,6,$ by the Order Test below, with $,color{#0a0}{p = 2,3}$



                Order Test $ ,a,$ has order $,n iff a^{large n} = 1 $ but $ color{#0a0}{a^{large n/p} neq 1},$ for every prime $,pmid n$



                Proof $, (Leftarrow), $ Let $,a,$ have $,color{#c00}{{rm order} k}.,$ Then $,kmid n,$ (proof). $ $ If $:k < n,$ then $,k,$ is proper divisor of $,n,$ so by unique factorization $,k,$ arises by deleting at least one prime $,p,$ from the prime factorization of $,n,,$ so $,kmid n/p,,$ say $, kj = n/p, $ so $ color{#0a0}{a^{large n/p}} = (color{#c00}{a^{large k}})^{large j} = color{#c00}1^{large j} = color{#0a0}1,$ contra $rmcolor{#0a0}{hypothesis}$. So $,k=n.$ $, (Rightarrow) $ Clear






                share|cite|improve this answer











                $endgroup$


















                  4












                  $begingroup$

                  $!bmod p!:, a^3equiv 1, anotequiv 1,$ so $,0equiv a^3-1equiv (a-1)(a^2+a+1),Rightarrow,a^2+a+1equiv 0$



                  Thus $ 1+aequiv -a^2. $ To compute $,(1+a)^n,$ we power this congruence:



                  Thus $,(1+a)^6equiv, a^{12}equiv 1, $ by $,a^3equiv 1$



                  But $ color{#0a0}{(1+a)^2}equiv a^4, equiv, a, color{#0a0}{notequiv 1},$ by hypothesis



                  and $ , color{#0a0}{(1+a)^3}!equiv! -a^6!equiv! -1 color{#0a0}{notequiv 1},$ by $,pneq 2, $ (where only order $1$ is possible)



                  Thus $,1+a,$ has order $,6,$ by the Order Test below, with $,color{#0a0}{p = 2,3}$



                  Order Test $ ,a,$ has order $,n iff a^{large n} = 1 $ but $ color{#0a0}{a^{large n/p} neq 1},$ for every prime $,pmid n$



                  Proof $, (Leftarrow), $ Let $,a,$ have $,color{#c00}{{rm order} k}.,$ Then $,kmid n,$ (proof). $ $ If $:k < n,$ then $,k,$ is proper divisor of $,n,$ so by unique factorization $,k,$ arises by deleting at least one prime $,p,$ from the prime factorization of $,n,,$ so $,kmid n/p,,$ say $, kj = n/p, $ so $ color{#0a0}{a^{large n/p}} = (color{#c00}{a^{large k}})^{large j} = color{#c00}1^{large j} = color{#0a0}1,$ contra $rmcolor{#0a0}{hypothesis}$. So $,k=n.$ $, (Rightarrow) $ Clear






                  share|cite|improve this answer











                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    $!bmod p!:, a^3equiv 1, anotequiv 1,$ so $,0equiv a^3-1equiv (a-1)(a^2+a+1),Rightarrow,a^2+a+1equiv 0$



                    Thus $ 1+aequiv -a^2. $ To compute $,(1+a)^n,$ we power this congruence:



                    Thus $,(1+a)^6equiv, a^{12}equiv 1, $ by $,a^3equiv 1$



                    But $ color{#0a0}{(1+a)^2}equiv a^4, equiv, a, color{#0a0}{notequiv 1},$ by hypothesis



                    and $ , color{#0a0}{(1+a)^3}!equiv! -a^6!equiv! -1 color{#0a0}{notequiv 1},$ by $,pneq 2, $ (where only order $1$ is possible)



                    Thus $,1+a,$ has order $,6,$ by the Order Test below, with $,color{#0a0}{p = 2,3}$



                    Order Test $ ,a,$ has order $,n iff a^{large n} = 1 $ but $ color{#0a0}{a^{large n/p} neq 1},$ for every prime $,pmid n$



                    Proof $, (Leftarrow), $ Let $,a,$ have $,color{#c00}{{rm order} k}.,$ Then $,kmid n,$ (proof). $ $ If $:k < n,$ then $,k,$ is proper divisor of $,n,$ so by unique factorization $,k,$ arises by deleting at least one prime $,p,$ from the prime factorization of $,n,,$ so $,kmid n/p,,$ say $, kj = n/p, $ so $ color{#0a0}{a^{large n/p}} = (color{#c00}{a^{large k}})^{large j} = color{#c00}1^{large j} = color{#0a0}1,$ contra $rmcolor{#0a0}{hypothesis}$. So $,k=n.$ $, (Rightarrow) $ Clear






                    share|cite|improve this answer











                    $endgroup$



                    $!bmod p!:, a^3equiv 1, anotequiv 1,$ so $,0equiv a^3-1equiv (a-1)(a^2+a+1),Rightarrow,a^2+a+1equiv 0$



                    Thus $ 1+aequiv -a^2. $ To compute $,(1+a)^n,$ we power this congruence:



                    Thus $,(1+a)^6equiv, a^{12}equiv 1, $ by $,a^3equiv 1$



                    But $ color{#0a0}{(1+a)^2}equiv a^4, equiv, a, color{#0a0}{notequiv 1},$ by hypothesis



                    and $ , color{#0a0}{(1+a)^3}!equiv! -a^6!equiv! -1 color{#0a0}{notequiv 1},$ by $,pneq 2, $ (where only order $1$ is possible)



                    Thus $,1+a,$ has order $,6,$ by the Order Test below, with $,color{#0a0}{p = 2,3}$



                    Order Test $ ,a,$ has order $,n iff a^{large n} = 1 $ but $ color{#0a0}{a^{large n/p} neq 1},$ for every prime $,pmid n$



                    Proof $, (Leftarrow), $ Let $,a,$ have $,color{#c00}{{rm order} k}.,$ Then $,kmid n,$ (proof). $ $ If $:k < n,$ then $,k,$ is proper divisor of $,n,$ so by unique factorization $,k,$ arises by deleting at least one prime $,p,$ from the prime factorization of $,n,,$ so $,kmid n/p,,$ say $, kj = n/p, $ so $ color{#0a0}{a^{large n/p}} = (color{#c00}{a^{large k}})^{large j} = color{#c00}1^{large j} = color{#0a0}1,$ contra $rmcolor{#0a0}{hypothesis}$. So $,k=n.$ $, (Rightarrow) $ Clear







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 28 '18 at 18:06

























                    answered Dec 28 '18 at 17:25









                    Bill DubuqueBill Dubuque

                    212k29195654




                    212k29195654























                        3












                        $begingroup$

                        As $a$ has order $3$, $a^3equiv 1 pmod p$. Hence $a^3-1 equiv 0 pmod p$.
                        So $a^3-1=(a-1)(1+a+a^2)equiv 0 pmod p$.



                        Now, $(1+a)^2=1+2a+a^2$. This will be a nice hint. Use the previous part in some way.






                        share|cite|improve this answer









                        $endgroup$


















                          3












                          $begingroup$

                          As $a$ has order $3$, $a^3equiv 1 pmod p$. Hence $a^3-1 equiv 0 pmod p$.
                          So $a^3-1=(a-1)(1+a+a^2)equiv 0 pmod p$.



                          Now, $(1+a)^2=1+2a+a^2$. This will be a nice hint. Use the previous part in some way.






                          share|cite|improve this answer









                          $endgroup$
















                            3












                            3








                            3





                            $begingroup$

                            As $a$ has order $3$, $a^3equiv 1 pmod p$. Hence $a^3-1 equiv 0 pmod p$.
                            So $a^3-1=(a-1)(1+a+a^2)equiv 0 pmod p$.



                            Now, $(1+a)^2=1+2a+a^2$. This will be a nice hint. Use the previous part in some way.






                            share|cite|improve this answer









                            $endgroup$



                            As $a$ has order $3$, $a^3equiv 1 pmod p$. Hence $a^3-1 equiv 0 pmod p$.
                            So $a^3-1=(a-1)(1+a+a^2)equiv 0 pmod p$.



                            Now, $(1+a)^2=1+2a+a^2$. This will be a nice hint. Use the previous part in some way.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 28 '18 at 17:05









                            toric_actionstoric_actions

                            1088




                            1088






























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