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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