Showing $gcd(2^m-1,2^n+1)=1$












7












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A student of mine has been self-studying some elementary number theory. She came by my office today and asked if I had any hints on how to prove the statement




If $m$ is odd then $gcd(2^m-1,2^n+1)=1$.




It's been a while since I took number theory and I'm not sure what to do. She said she is learning about congruences, primitive roots, and power residues. She has not taken any group theory.






elementary-number-theory divisibility greatest-common-divisor







share|cite|improve this question











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    7












    $begingroup$


    A student of mine has been self-studying some elementary number theory. She came by my office today and asked if I had any hints on how to prove the statement




    If $m$ is odd then $gcd(2^m-1,2^n+1)=1$.




    It's been a while since I took number theory and I'm not sure what to do. She said she is learning about congruences, primitive roots, and power residues. She has not taken any group theory.






    elementary-number-theory divisibility greatest-common-divisor







    share|cite|improve this question











    $endgroup$















      7












      7








      7


      4



      $begingroup$


      A student of mine has been self-studying some elementary number theory. She came by my office today and asked if I had any hints on how to prove the statement




      If $m$ is odd then $gcd(2^m-1,2^n+1)=1$.




      It's been a while since I took number theory and I'm not sure what to do. She said she is learning about congruences, primitive roots, and power residues. She has not taken any group theory.






      elementary-number-theory divisibility greatest-common-divisor







      share|cite|improve this question











      $endgroup$




      A student of mine has been self-studying some elementary number theory. She came by my office today and asked if I had any hints on how to prove the statement




      If $m$ is odd then $gcd(2^m-1,2^n+1)=1$.




      It's been a while since I took number theory and I'm not sure what to do. She said she is learning about congruences, primitive roots, and power residues. She has not taken any group theory.






      elementary-number-theory divisibility greatest-common-divisor




      elementary-number-theory divisibility greatest-common-divisor






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 16 '16 at 21:37









      Martin Sleziak

      44.7k8117272




      44.7k8117272










      asked Nov 18 '13 at 19:06









      Joe Johnson 126Joe Johnson 126

      13.8k32770




      13.8k32770






















          1 Answer
          1






          active

          oldest

          votes


















          9












          $begingroup$

          If an odd prime $p$ divides $2^n+1$, then the order of $2$ modulo $p$ is even (it is a divisor of $2n$, but not of $n$). If an odd prime $q$ divides $2^m-1$ with $m$ odd, then the order of $2$ modulo $q$ is odd (it is a divisor of $m$). Hence $p neq q$. Since $2^m - 1$ is odd for $m > 0$, in particular all odd $m$, the greatest common divisor cannot be even. So no prime divides both, $2^n+1$ and $2^m-1$.



          Alternatively, we can use



          $$gcd (2^t-1, 2^u-1) = 2^{gcd (t,u)}-1tag{1}$$



          to conclude



          $$gcd (2^m-1, 2^{2n}-1) = 2^{gcd(m,2n)}-1.$$



          But since $m$ is odd, we have $gcd (m,2n) = gcd(m,n)$, and hence



          $$2^{gcd(m,2n)}-1 mid 2^n-1,$$



          which, since



          $$gcd(2^n-1,2^n+1) = gcd(2^n-1,2) mid 2$$



          and $2^{gcd(m,2n)}-1$ is odd, implies $gcd (2^{gcd(m,2n)}-1,2^n+1) = 1$ and hence $gcd(2^m-1,2^n+1) = 1$.



          To see $(1)$, write $u = qcdot t + r$ with $0 leqslant r < t$, and



          $$2^u-1 = 2^rleft(2^{qcdot t}-1right) + left(2^r-1right),$$



          which, since $2^t-1 mid (2^t)^q-1$, yields



          $$gcd(2^t-1,2^u-1) = gcd(2^t-1,2^r-1),$$



          and continuing the Euclidean algorithm for the exponents finally yields $(1)$.






          share|cite|improve this answer











          $endgroup$













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            1 Answer
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            active

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            9












            $begingroup$

            If an odd prime $p$ divides $2^n+1$, then the order of $2$ modulo $p$ is even (it is a divisor of $2n$, but not of $n$). If an odd prime $q$ divides $2^m-1$ with $m$ odd, then the order of $2$ modulo $q$ is odd (it is a divisor of $m$). Hence $p neq q$. Since $2^m - 1$ is odd for $m > 0$, in particular all odd $m$, the greatest common divisor cannot be even. So no prime divides both, $2^n+1$ and $2^m-1$.



            Alternatively, we can use



            $$gcd (2^t-1, 2^u-1) = 2^{gcd (t,u)}-1tag{1}$$



            to conclude



            $$gcd (2^m-1, 2^{2n}-1) = 2^{gcd(m,2n)}-1.$$



            But since $m$ is odd, we have $gcd (m,2n) = gcd(m,n)$, and hence



            $$2^{gcd(m,2n)}-1 mid 2^n-1,$$



            which, since



            $$gcd(2^n-1,2^n+1) = gcd(2^n-1,2) mid 2$$



            and $2^{gcd(m,2n)}-1$ is odd, implies $gcd (2^{gcd(m,2n)}-1,2^n+1) = 1$ and hence $gcd(2^m-1,2^n+1) = 1$.



            To see $(1)$, write $u = qcdot t + r$ with $0 leqslant r < t$, and



            $$2^u-1 = 2^rleft(2^{qcdot t}-1right) + left(2^r-1right),$$



            which, since $2^t-1 mid (2^t)^q-1$, yields



            $$gcd(2^t-1,2^u-1) = gcd(2^t-1,2^r-1),$$



            and continuing the Euclidean algorithm for the exponents finally yields $(1)$.






            share|cite|improve this answer











            $endgroup$


















              9












              $begingroup$

              If an odd prime $p$ divides $2^n+1$, then the order of $2$ modulo $p$ is even (it is a divisor of $2n$, but not of $n$). If an odd prime $q$ divides $2^m-1$ with $m$ odd, then the order of $2$ modulo $q$ is odd (it is a divisor of $m$). Hence $p neq q$. Since $2^m - 1$ is odd for $m > 0$, in particular all odd $m$, the greatest common divisor cannot be even. So no prime divides both, $2^n+1$ and $2^m-1$.



              Alternatively, we can use



              $$gcd (2^t-1, 2^u-1) = 2^{gcd (t,u)}-1tag{1}$$



              to conclude



              $$gcd (2^m-1, 2^{2n}-1) = 2^{gcd(m,2n)}-1.$$



              But since $m$ is odd, we have $gcd (m,2n) = gcd(m,n)$, and hence



              $$2^{gcd(m,2n)}-1 mid 2^n-1,$$



              which, since



              $$gcd(2^n-1,2^n+1) = gcd(2^n-1,2) mid 2$$



              and $2^{gcd(m,2n)}-1$ is odd, implies $gcd (2^{gcd(m,2n)}-1,2^n+1) = 1$ and hence $gcd(2^m-1,2^n+1) = 1$.



              To see $(1)$, write $u = qcdot t + r$ with $0 leqslant r < t$, and



              $$2^u-1 = 2^rleft(2^{qcdot t}-1right) + left(2^r-1right),$$



              which, since $2^t-1 mid (2^t)^q-1$, yields



              $$gcd(2^t-1,2^u-1) = gcd(2^t-1,2^r-1),$$



              and continuing the Euclidean algorithm for the exponents finally yields $(1)$.






              share|cite|improve this answer











              $endgroup$
















                9












                9








                9





                $begingroup$

                If an odd prime $p$ divides $2^n+1$, then the order of $2$ modulo $p$ is even (it is a divisor of $2n$, but not of $n$). If an odd prime $q$ divides $2^m-1$ with $m$ odd, then the order of $2$ modulo $q$ is odd (it is a divisor of $m$). Hence $p neq q$. Since $2^m - 1$ is odd for $m > 0$, in particular all odd $m$, the greatest common divisor cannot be even. So no prime divides both, $2^n+1$ and $2^m-1$.



                Alternatively, we can use



                $$gcd (2^t-1, 2^u-1) = 2^{gcd (t,u)}-1tag{1}$$



                to conclude



                $$gcd (2^m-1, 2^{2n}-1) = 2^{gcd(m,2n)}-1.$$



                But since $m$ is odd, we have $gcd (m,2n) = gcd(m,n)$, and hence



                $$2^{gcd(m,2n)}-1 mid 2^n-1,$$



                which, since



                $$gcd(2^n-1,2^n+1) = gcd(2^n-1,2) mid 2$$



                and $2^{gcd(m,2n)}-1$ is odd, implies $gcd (2^{gcd(m,2n)}-1,2^n+1) = 1$ and hence $gcd(2^m-1,2^n+1) = 1$.



                To see $(1)$, write $u = qcdot t + r$ with $0 leqslant r < t$, and



                $$2^u-1 = 2^rleft(2^{qcdot t}-1right) + left(2^r-1right),$$



                which, since $2^t-1 mid (2^t)^q-1$, yields



                $$gcd(2^t-1,2^u-1) = gcd(2^t-1,2^r-1),$$



                and continuing the Euclidean algorithm for the exponents finally yields $(1)$.






                share|cite|improve this answer











                $endgroup$



                If an odd prime $p$ divides $2^n+1$, then the order of $2$ modulo $p$ is even (it is a divisor of $2n$, but not of $n$). If an odd prime $q$ divides $2^m-1$ with $m$ odd, then the order of $2$ modulo $q$ is odd (it is a divisor of $m$). Hence $p neq q$. Since $2^m - 1$ is odd for $m > 0$, in particular all odd $m$, the greatest common divisor cannot be even. So no prime divides both, $2^n+1$ and $2^m-1$.



                Alternatively, we can use



                $$gcd (2^t-1, 2^u-1) = 2^{gcd (t,u)}-1tag{1}$$



                to conclude



                $$gcd (2^m-1, 2^{2n}-1) = 2^{gcd(m,2n)}-1.$$



                But since $m$ is odd, we have $gcd (m,2n) = gcd(m,n)$, and hence



                $$2^{gcd(m,2n)}-1 mid 2^n-1,$$



                which, since



                $$gcd(2^n-1,2^n+1) = gcd(2^n-1,2) mid 2$$



                and $2^{gcd(m,2n)}-1$ is odd, implies $gcd (2^{gcd(m,2n)}-1,2^n+1) = 1$ and hence $gcd(2^m-1,2^n+1) = 1$.



                To see $(1)$, write $u = qcdot t + r$ with $0 leqslant r < t$, and



                $$2^u-1 = 2^rleft(2^{qcdot t}-1right) + left(2^r-1right),$$



                which, since $2^t-1 mid (2^t)^q-1$, yields



                $$gcd(2^t-1,2^u-1) = gcd(2^t-1,2^r-1),$$



                and continuing the Euclidean algorithm for the exponents finally yields $(1)$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 18 '13 at 22:11

























                answered Nov 18 '13 at 19:09









                Daniel Fischer♦Daniel Fischer

                173k16161283




                173k16161283






























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