Find the Primitive Roots $mod 31$












2












$begingroup$


My approach:



There exist $phi(31-1) = phi(30) = 8$ primitive roots.



If $x^6 notequiv 1$,$x^{10} notequiv 1$, and $x^{15} notequiv 1$, then $x$ is a primitive root modulo $31$.



$x = 1, 2$ fails this but $x = 3$ passes this, thus $3$ is a primitive root.



I then know that ${3^0, 3^1, 3^2, dots, 3^{29}}$ is a residue system mod $31$.



How can I then determine which elements are the primitive roots of this set?






elementary-number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Well, $g=3^2$ isn't a primitive root because $gcd(2,30)=2$ and $g^{15}=1$, noting that $15=frac {30}2$. Do you see the pattern?
    $endgroup$
    – lulu
    2 days ago










  • $begingroup$
    Phrased differently, you say that you know that there are $varphi(30)=8$ primitive roots. How do you know that? The proof of that tells you how to find all the others, given one.
    $endgroup$
    – lulu
    2 days ago
















2












$begingroup$


My approach:



There exist $phi(31-1) = phi(30) = 8$ primitive roots.



If $x^6 notequiv 1$,$x^{10} notequiv 1$, and $x^{15} notequiv 1$, then $x$ is a primitive root modulo $31$.



$x = 1, 2$ fails this but $x = 3$ passes this, thus $3$ is a primitive root.



I then know that ${3^0, 3^1, 3^2, dots, 3^{29}}$ is a residue system mod $31$.



How can I then determine which elements are the primitive roots of this set?






elementary-number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Well, $g=3^2$ isn't a primitive root because $gcd(2,30)=2$ and $g^{15}=1$, noting that $15=frac {30}2$. Do you see the pattern?
    $endgroup$
    – lulu
    2 days ago










  • $begingroup$
    Phrased differently, you say that you know that there are $varphi(30)=8$ primitive roots. How do you know that? The proof of that tells you how to find all the others, given one.
    $endgroup$
    – lulu
    2 days ago














2












2








2





$begingroup$


My approach:



There exist $phi(31-1) = phi(30) = 8$ primitive roots.



If $x^6 notequiv 1$,$x^{10} notequiv 1$, and $x^{15} notequiv 1$, then $x$ is a primitive root modulo $31$.



$x = 1, 2$ fails this but $x = 3$ passes this, thus $3$ is a primitive root.



I then know that ${3^0, 3^1, 3^2, dots, 3^{29}}$ is a residue system mod $31$.



How can I then determine which elements are the primitive roots of this set?






elementary-number-theory







share|cite|improve this question











$endgroup$




My approach:



There exist $phi(31-1) = phi(30) = 8$ primitive roots.



If $x^6 notequiv 1$,$x^{10} notequiv 1$, and $x^{15} notequiv 1$, then $x$ is a primitive root modulo $31$.



$x = 1, 2$ fails this but $x = 3$ passes this, thus $3$ is a primitive root.



I then know that ${3^0, 3^1, 3^2, dots, 3^{29}}$ is a residue system mod $31$.



How can I then determine which elements are the primitive roots of this set?






elementary-number-theory




elementary-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Asaf Karagila♦

307k33439770




307k33439770










asked 2 days ago









Bryden CBryden C

31919




31919












  • $begingroup$
    Well, $g=3^2$ isn't a primitive root because $gcd(2,30)=2$ and $g^{15}=1$, noting that $15=frac {30}2$. Do you see the pattern?
    $endgroup$
    – lulu
    2 days ago










  • $begingroup$
    Phrased differently, you say that you know that there are $varphi(30)=8$ primitive roots. How do you know that? The proof of that tells you how to find all the others, given one.
    $endgroup$
    – lulu
    2 days ago


















  • $begingroup$
    Well, $g=3^2$ isn't a primitive root because $gcd(2,30)=2$ and $g^{15}=1$, noting that $15=frac {30}2$. Do you see the pattern?
    $endgroup$
    – lulu
    2 days ago










  • $begingroup$
    Phrased differently, you say that you know that there are $varphi(30)=8$ primitive roots. How do you know that? The proof of that tells you how to find all the others, given one.
    $endgroup$
    – lulu
    2 days ago
















$begingroup$
Well, $g=3^2$ isn't a primitive root because $gcd(2,30)=2$ and $g^{15}=1$, noting that $15=frac {30}2$. Do you see the pattern?
$endgroup$
– lulu
2 days ago




$begingroup$
Well, $g=3^2$ isn't a primitive root because $gcd(2,30)=2$ and $g^{15}=1$, noting that $15=frac {30}2$. Do you see the pattern?
$endgroup$
– lulu
2 days ago












$begingroup$
Phrased differently, you say that you know that there are $varphi(30)=8$ primitive roots. How do you know that? The proof of that tells you how to find all the others, given one.
$endgroup$
– lulu
2 days ago




$begingroup$
Phrased differently, you say that you know that there are $varphi(30)=8$ primitive roots. How do you know that? The proof of that tells you how to find all the others, given one.
$endgroup$
– lulu
2 days ago










3 Answers
3






active

oldest

votes


















3












$begingroup$

There are indeed $phi(phi (31))=8$ primitive roots modulo $31$ and you can find them as described here:



Finding a primitive root of a prime number



For example, $3^kequiv 1bmod 31$ only holds for $k=30$, if $1le kle 30$. Hence $3$ is a primitive root modulo $31$. Now compute the orders of powers of $3$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Once you found one primitive root, the others are its powers which are relatively prime to $phi(31)=30$. The numbers in ${0,1,2,...,29}$ which are relatively prime to $30$ are $1,7,11,13,17,19,23,29$ and hence the primitive roots are $3,3^7,3^{11},...,3^{29}$.



    The reason why this is the case is the general formula $ord_n(a^k)=frac{ord_n(a)}{gcd(k,ord_n(a))}$.






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$


      I then know that ${3^0,3^1,3^2,…,3^{29}}$ is a residue system $mod 31$.




      And you are sooo close.



      $(3^k)^m = 3^{mk}$. So for $3^k$ to be a primitive root we need $mk$ to not be a multiple of $30$ for any natural $m < 30$.



      In other words if $k$ is relatively prime to $30$.



      In fact, that is precisely why we know there are $phi(30)$ primitive roots.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        There are indeed $phi(phi (31))=8$ primitive roots modulo $31$ and you can find them as described here:



        Finding a primitive root of a prime number



        For example, $3^kequiv 1bmod 31$ only holds for $k=30$, if $1le kle 30$. Hence $3$ is a primitive root modulo $31$. Now compute the orders of powers of $3$.






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          There are indeed $phi(phi (31))=8$ primitive roots modulo $31$ and you can find them as described here:



          Finding a primitive root of a prime number



          For example, $3^kequiv 1bmod 31$ only holds for $k=30$, if $1le kle 30$. Hence $3$ is a primitive root modulo $31$. Now compute the orders of powers of $3$.






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            There are indeed $phi(phi (31))=8$ primitive roots modulo $31$ and you can find them as described here:



            Finding a primitive root of a prime number



            For example, $3^kequiv 1bmod 31$ only holds for $k=30$, if $1le kle 30$. Hence $3$ is a primitive root modulo $31$. Now compute the orders of powers of $3$.






            share|cite|improve this answer









            $endgroup$



            There are indeed $phi(phi (31))=8$ primitive roots modulo $31$ and you can find them as described here:



            Finding a primitive root of a prime number



            For example, $3^kequiv 1bmod 31$ only holds for $k=30$, if $1le kle 30$. Hence $3$ is a primitive root modulo $31$. Now compute the orders of powers of $3$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 days ago









            Dietrich BurdeDietrich Burde

            81.3k648106




            81.3k648106























                2












                $begingroup$

                Once you found one primitive root, the others are its powers which are relatively prime to $phi(31)=30$. The numbers in ${0,1,2,...,29}$ which are relatively prime to $30$ are $1,7,11,13,17,19,23,29$ and hence the primitive roots are $3,3^7,3^{11},...,3^{29}$.



                The reason why this is the case is the general formula $ord_n(a^k)=frac{ord_n(a)}{gcd(k,ord_n(a))}$.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  Once you found one primitive root, the others are its powers which are relatively prime to $phi(31)=30$. The numbers in ${0,1,2,...,29}$ which are relatively prime to $30$ are $1,7,11,13,17,19,23,29$ and hence the primitive roots are $3,3^7,3^{11},...,3^{29}$.



                  The reason why this is the case is the general formula $ord_n(a^k)=frac{ord_n(a)}{gcd(k,ord_n(a))}$.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Once you found one primitive root, the others are its powers which are relatively prime to $phi(31)=30$. The numbers in ${0,1,2,...,29}$ which are relatively prime to $30$ are $1,7,11,13,17,19,23,29$ and hence the primitive roots are $3,3^7,3^{11},...,3^{29}$.



                    The reason why this is the case is the general formula $ord_n(a^k)=frac{ord_n(a)}{gcd(k,ord_n(a))}$.






                    share|cite|improve this answer









                    $endgroup$



                    Once you found one primitive root, the others are its powers which are relatively prime to $phi(31)=30$. The numbers in ${0,1,2,...,29}$ which are relatively prime to $30$ are $1,7,11,13,17,19,23,29$ and hence the primitive roots are $3,3^7,3^{11},...,3^{29}$.



                    The reason why this is the case is the general formula $ord_n(a^k)=frac{ord_n(a)}{gcd(k,ord_n(a))}$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 days ago









                    MarkMark

                    10.4k1622




                    10.4k1622























                        2












                        $begingroup$


                        I then know that ${3^0,3^1,3^2,…,3^{29}}$ is a residue system $mod 31$.




                        And you are sooo close.



                        $(3^k)^m = 3^{mk}$. So for $3^k$ to be a primitive root we need $mk$ to not be a multiple of $30$ for any natural $m < 30$.



                        In other words if $k$ is relatively prime to $30$.



                        In fact, that is precisely why we know there are $phi(30)$ primitive roots.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$


                          I then know that ${3^0,3^1,3^2,…,3^{29}}$ is a residue system $mod 31$.




                          And you are sooo close.



                          $(3^k)^m = 3^{mk}$. So for $3^k$ to be a primitive root we need $mk$ to not be a multiple of $30$ for any natural $m < 30$.



                          In other words if $k$ is relatively prime to $30$.



                          In fact, that is precisely why we know there are $phi(30)$ primitive roots.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$


                            I then know that ${3^0,3^1,3^2,…,3^{29}}$ is a residue system $mod 31$.




                            And you are sooo close.



                            $(3^k)^m = 3^{mk}$. So for $3^k$ to be a primitive root we need $mk$ to not be a multiple of $30$ for any natural $m < 30$.



                            In other words if $k$ is relatively prime to $30$.



                            In fact, that is precisely why we know there are $phi(30)$ primitive roots.






                            share|cite|improve this answer









                            $endgroup$




                            I then know that ${3^0,3^1,3^2,…,3^{29}}$ is a residue system $mod 31$.




                            And you are sooo close.



                            $(3^k)^m = 3^{mk}$. So for $3^k$ to be a primitive root we need $mk$ to not be a multiple of $30$ for any natural $m < 30$.



                            In other words if $k$ is relatively prime to $30$.



                            In fact, that is precisely why we know there are $phi(30)$ primitive roots.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 days ago









                            fleabloodfleablood

                            73.4k22791




                            73.4k22791






























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                                SlayerMusikgrupper bildade 1981Amerikanska thrash metal-grupper amerikansktthrash metalHuntington Park1981gitarristernaJeff HannemanKerry KingbasistenTom ArayatrummisenDave LombardoGary HoltPaul Bostaphcertifierats med guldMetallicaAnthraxMegadeththrash metal1986extrema metalgenrerdeath metal22 januari2018USAKanadaEuropaHuntington ParkKalifornien1981gitarristernaJeff HannemanKerry KingbassångDave Lombardoheavy metalbrittiskaIron MaidenJudas PriestthrashMetallicaRattSteeler1982demoundergroundMetal Blade1983high schoolGene HoglanDeathStrapping Young Ladkassettvinyl1984LondonMarquee Club1984Eldorado StudiosTrack Record StudiosLos Angelesmastrade1985CDCombat Recordsextreme metalVenomExodusStudio 54 club1985VHSCombat RecordsBelgienUFORick RubinDef Jam Recordship hoprockakterThe Black CrowesRun–D.M.C.LL Cool JBeastie BoysthrashbpmAngel of DeathRaining BloodAndy WallaceHowie Weinberg1986bildskivaBillboard 200RIAAMetal HammerKerrang!Angela Gossowdeath metalArch Enemythrash metalOverk...
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