Elementary number theory proof difficulty












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I am having trouble proving this implication involving greatest common divisors



Prove that for all $a,binBbb{Z}$, if $gcd(a,b)=1$, then $gcd(a^n,b^m)=1$ for all $n,minBbb{N}$



Since $gcd(a,b)=1$, then $exists x,yin Bbb{Z}$ such that $ax+by=1$ by the Coprimeness Characterization Theorem.



Raising both sides to the power to the power of $n$ gives $(ax+by)^n=1$



$$sum^n_{k=0}{n choose k}(ax)^{n-k}(by)^k=1implies a^nx^n+bsum^n_{k=1}{n choose k}(ax)^{n-k}b^{k-1}y^k=1$$



A similar argument of course applies for raising both sides to the power of $m$: $(ax+by)^m=1$



$$sum^m_{k=0}{m choose k}(ax)^k(by)^{m-k}=1implies b^my^m+asum^m_{k=1}{m choose k}a^{k-1}x^k(by)^{m-k}=1$$



From both of these I can conclude separately that $gcd(a^n,b)=1$ and $gcd(a,b^m)=1$, but I don't know how to prove that $gcd(a^n,b^m)=1$ with $n$ and $m$ combined. Any help?






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    $begingroup$


    I am having trouble proving this implication involving greatest common divisors



    Prove that for all $a,binBbb{Z}$, if $gcd(a,b)=1$, then $gcd(a^n,b^m)=1$ for all $n,minBbb{N}$



    Since $gcd(a,b)=1$, then $exists x,yin Bbb{Z}$ such that $ax+by=1$ by the Coprimeness Characterization Theorem.



    Raising both sides to the power to the power of $n$ gives $(ax+by)^n=1$



    $$sum^n_{k=0}{n choose k}(ax)^{n-k}(by)^k=1implies a^nx^n+bsum^n_{k=1}{n choose k}(ax)^{n-k}b^{k-1}y^k=1$$



    A similar argument of course applies for raising both sides to the power of $m$: $(ax+by)^m=1$



    $$sum^m_{k=0}{m choose k}(ax)^k(by)^{m-k}=1implies b^my^m+asum^m_{k=1}{m choose k}a^{k-1}x^k(by)^{m-k}=1$$



    From both of these I can conclude separately that $gcd(a^n,b)=1$ and $gcd(a,b^m)=1$, but I don't know how to prove that $gcd(a^n,b^m)=1$ with $n$ and $m$ combined. Any help?






    elementary-number-theory







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      $begingroup$


      I am having trouble proving this implication involving greatest common divisors



      Prove that for all $a,binBbb{Z}$, if $gcd(a,b)=1$, then $gcd(a^n,b^m)=1$ for all $n,minBbb{N}$



      Since $gcd(a,b)=1$, then $exists x,yin Bbb{Z}$ such that $ax+by=1$ by the Coprimeness Characterization Theorem.



      Raising both sides to the power to the power of $n$ gives $(ax+by)^n=1$



      $$sum^n_{k=0}{n choose k}(ax)^{n-k}(by)^k=1implies a^nx^n+bsum^n_{k=1}{n choose k}(ax)^{n-k}b^{k-1}y^k=1$$



      A similar argument of course applies for raising both sides to the power of $m$: $(ax+by)^m=1$



      $$sum^m_{k=0}{m choose k}(ax)^k(by)^{m-k}=1implies b^my^m+asum^m_{k=1}{m choose k}a^{k-1}x^k(by)^{m-k}=1$$



      From both of these I can conclude separately that $gcd(a^n,b)=1$ and $gcd(a,b^m)=1$, but I don't know how to prove that $gcd(a^n,b^m)=1$ with $n$ and $m$ combined. Any help?






      elementary-number-theory







      share|cite|improve this question









      $endgroup$




      I am having trouble proving this implication involving greatest common divisors



      Prove that for all $a,binBbb{Z}$, if $gcd(a,b)=1$, then $gcd(a^n,b^m)=1$ for all $n,minBbb{N}$



      Since $gcd(a,b)=1$, then $exists x,yin Bbb{Z}$ such that $ax+by=1$ by the Coprimeness Characterization Theorem.



      Raising both sides to the power to the power of $n$ gives $(ax+by)^n=1$



      $$sum^n_{k=0}{n choose k}(ax)^{n-k}(by)^k=1implies a^nx^n+bsum^n_{k=1}{n choose k}(ax)^{n-k}b^{k-1}y^k=1$$



      A similar argument of course applies for raising both sides to the power of $m$: $(ax+by)^m=1$



      $$sum^m_{k=0}{m choose k}(ax)^k(by)^{m-k}=1implies b^my^m+asum^m_{k=1}{m choose k}a^{k-1}x^k(by)^{m-k}=1$$



      From both of these I can conclude separately that $gcd(a^n,b)=1$ and $gcd(a,b^m)=1$, but I don't know how to prove that $gcd(a^n,b^m)=1$ with $n$ and $m$ combined. Any help?






      elementary-number-theory




      elementary-number-theory






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      asked Dec 6 '18 at 22:16









      Anson PangAnson Pang

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          $begingroup$

          An alternate and arguably a better approach is as follows: Suppose the $gcd(a^m,b^n)>1$. Let a prime $p mid gcd(a^m,b^n)$. Then $p mid a^m$ and $p mid b^n$. By the prime property ($p$ divides a product, hence divides at least one component of the product), we get $p mid a$ and $p mid b$. Thus $p mid gcd(a,b)$. But $gcd(a,b)=1$. So we get a contradiction.






          share|cite|improve this answer









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          • $begingroup$
            How did you go from p dividing gcd(a^m,b^n) to p dividing a^m and p dividing b^n
            $endgroup$
            – Anson Pang
            Dec 6 '18 at 23:44










          • $begingroup$
            @AnsonPang Let $d=gcd(a^m,b^n)$. Then by the definition of $gcd$, $d mid a^m$ and $d mid b^n$. If $p mid d$, then $p$ is a common divisor as well.
            $endgroup$
            – Anurag A
            Dec 7 '18 at 0:08





















          0












          $begingroup$

          Raise the equation to the power $m+n-1$ (in the exceptional case $m=n=0$, it is clear that $gcd(1, 1) = 1$):
          $$sum_{i=0}^{m+n-1} binom{m+n-1}{i} a^i x^i b^{m+n-1-i} y^{m+n-1-i} = 1.$$
          Now each term is divisible either by $a^m$, if $i ge m$, or by $b^n$, if $m+n-1-i ge n$. (One of these must hold; otherwise, $i le m-1$ and $m+n-1-i le n-1$, and summing gives a contradiction $m+n-1 le m+n-2$.)






          share|cite|improve this answer









          $endgroup$





















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            That seems very complicated and I'd give up.



            If $gcd(a,b)= 1$ means that $a$ and $b$ have no prime factors in common. $a^n$ has only the prime factors that $a$ has and $b^m$ has only the prime factors that $b$ has.



            So $a^n$ and $b^n$ have no prime factors in common either. So $gcd(a^n, b^m) = 1$.



            Simple as that.






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              3 Answers
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              3 Answers
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              3












              $begingroup$

              An alternate and arguably a better approach is as follows: Suppose the $gcd(a^m,b^n)>1$. Let a prime $p mid gcd(a^m,b^n)$. Then $p mid a^m$ and $p mid b^n$. By the prime property ($p$ divides a product, hence divides at least one component of the product), we get $p mid a$ and $p mid b$. Thus $p mid gcd(a,b)$. But $gcd(a,b)=1$. So we get a contradiction.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                How did you go from p dividing gcd(a^m,b^n) to p dividing a^m and p dividing b^n
                $endgroup$
                – Anson Pang
                Dec 6 '18 at 23:44










              • $begingroup$
                @AnsonPang Let $d=gcd(a^m,b^n)$. Then by the definition of $gcd$, $d mid a^m$ and $d mid b^n$. If $p mid d$, then $p$ is a common divisor as well.
                $endgroup$
                – Anurag A
                Dec 7 '18 at 0:08


















              3












              $begingroup$

              An alternate and arguably a better approach is as follows: Suppose the $gcd(a^m,b^n)>1$. Let a prime $p mid gcd(a^m,b^n)$. Then $p mid a^m$ and $p mid b^n$. By the prime property ($p$ divides a product, hence divides at least one component of the product), we get $p mid a$ and $p mid b$. Thus $p mid gcd(a,b)$. But $gcd(a,b)=1$. So we get a contradiction.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                How did you go from p dividing gcd(a^m,b^n) to p dividing a^m and p dividing b^n
                $endgroup$
                – Anson Pang
                Dec 6 '18 at 23:44










              • $begingroup$
                @AnsonPang Let $d=gcd(a^m,b^n)$. Then by the definition of $gcd$, $d mid a^m$ and $d mid b^n$. If $p mid d$, then $p$ is a common divisor as well.
                $endgroup$
                – Anurag A
                Dec 7 '18 at 0:08
















              3












              3








              3





              $begingroup$

              An alternate and arguably a better approach is as follows: Suppose the $gcd(a^m,b^n)>1$. Let a prime $p mid gcd(a^m,b^n)$. Then $p mid a^m$ and $p mid b^n$. By the prime property ($p$ divides a product, hence divides at least one component of the product), we get $p mid a$ and $p mid b$. Thus $p mid gcd(a,b)$. But $gcd(a,b)=1$. So we get a contradiction.






              share|cite|improve this answer









              $endgroup$



              An alternate and arguably a better approach is as follows: Suppose the $gcd(a^m,b^n)>1$. Let a prime $p mid gcd(a^m,b^n)$. Then $p mid a^m$ and $p mid b^n$. By the prime property ($p$ divides a product, hence divides at least one component of the product), we get $p mid a$ and $p mid b$. Thus $p mid gcd(a,b)$. But $gcd(a,b)=1$. So we get a contradiction.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 6 '18 at 22:23









              Anurag AAnurag A

              26k12250




              26k12250












              • $begingroup$
                How did you go from p dividing gcd(a^m,b^n) to p dividing a^m and p dividing b^n
                $endgroup$
                – Anson Pang
                Dec 6 '18 at 23:44










              • $begingroup$
                @AnsonPang Let $d=gcd(a^m,b^n)$. Then by the definition of $gcd$, $d mid a^m$ and $d mid b^n$. If $p mid d$, then $p$ is a common divisor as well.
                $endgroup$
                – Anurag A
                Dec 7 '18 at 0:08




















              • $begingroup$
                How did you go from p dividing gcd(a^m,b^n) to p dividing a^m and p dividing b^n
                $endgroup$
                – Anson Pang
                Dec 6 '18 at 23:44










              • $begingroup$
                @AnsonPang Let $d=gcd(a^m,b^n)$. Then by the definition of $gcd$, $d mid a^m$ and $d mid b^n$. If $p mid d$, then $p$ is a common divisor as well.
                $endgroup$
                – Anurag A
                Dec 7 '18 at 0:08


















              $begingroup$
              How did you go from p dividing gcd(a^m,b^n) to p dividing a^m and p dividing b^n
              $endgroup$
              – Anson Pang
              Dec 6 '18 at 23:44




              $begingroup$
              How did you go from p dividing gcd(a^m,b^n) to p dividing a^m and p dividing b^n
              $endgroup$
              – Anson Pang
              Dec 6 '18 at 23:44












              $begingroup$
              @AnsonPang Let $d=gcd(a^m,b^n)$. Then by the definition of $gcd$, $d mid a^m$ and $d mid b^n$. If $p mid d$, then $p$ is a common divisor as well.
              $endgroup$
              – Anurag A
              Dec 7 '18 at 0:08






              $begingroup$
              @AnsonPang Let $d=gcd(a^m,b^n)$. Then by the definition of $gcd$, $d mid a^m$ and $d mid b^n$. If $p mid d$, then $p$ is a common divisor as well.
              $endgroup$
              – Anurag A
              Dec 7 '18 at 0:08













              0












              $begingroup$

              Raise the equation to the power $m+n-1$ (in the exceptional case $m=n=0$, it is clear that $gcd(1, 1) = 1$):
              $$sum_{i=0}^{m+n-1} binom{m+n-1}{i} a^i x^i b^{m+n-1-i} y^{m+n-1-i} = 1.$$
              Now each term is divisible either by $a^m$, if $i ge m$, or by $b^n$, if $m+n-1-i ge n$. (One of these must hold; otherwise, $i le m-1$ and $m+n-1-i le n-1$, and summing gives a contradiction $m+n-1 le m+n-2$.)






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Raise the equation to the power $m+n-1$ (in the exceptional case $m=n=0$, it is clear that $gcd(1, 1) = 1$):
                $$sum_{i=0}^{m+n-1} binom{m+n-1}{i} a^i x^i b^{m+n-1-i} y^{m+n-1-i} = 1.$$
                Now each term is divisible either by $a^m$, if $i ge m$, or by $b^n$, if $m+n-1-i ge n$. (One of these must hold; otherwise, $i le m-1$ and $m+n-1-i le n-1$, and summing gives a contradiction $m+n-1 le m+n-2$.)






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Raise the equation to the power $m+n-1$ (in the exceptional case $m=n=0$, it is clear that $gcd(1, 1) = 1$):
                  $$sum_{i=0}^{m+n-1} binom{m+n-1}{i} a^i x^i b^{m+n-1-i} y^{m+n-1-i} = 1.$$
                  Now each term is divisible either by $a^m$, if $i ge m$, or by $b^n$, if $m+n-1-i ge n$. (One of these must hold; otherwise, $i le m-1$ and $m+n-1-i le n-1$, and summing gives a contradiction $m+n-1 le m+n-2$.)






                  share|cite|improve this answer









                  $endgroup$



                  Raise the equation to the power $m+n-1$ (in the exceptional case $m=n=0$, it is clear that $gcd(1, 1) = 1$):
                  $$sum_{i=0}^{m+n-1} binom{m+n-1}{i} a^i x^i b^{m+n-1-i} y^{m+n-1-i} = 1.$$
                  Now each term is divisible either by $a^m$, if $i ge m$, or by $b^n$, if $m+n-1-i ge n$. (One of these must hold; otherwise, $i le m-1$ and $m+n-1-i le n-1$, and summing gives a contradiction $m+n-1 le m+n-2$.)







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 6 '18 at 22:33









                  Daniel ScheplerDaniel Schepler

                  8,7491620




                  8,7491620























                      0












                      $begingroup$

                      That seems very complicated and I'd give up.



                      If $gcd(a,b)= 1$ means that $a$ and $b$ have no prime factors in common. $a^n$ has only the prime factors that $a$ has and $b^m$ has only the prime factors that $b$ has.



                      So $a^n$ and $b^n$ have no prime factors in common either. So $gcd(a^n, b^m) = 1$.



                      Simple as that.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        That seems very complicated and I'd give up.



                        If $gcd(a,b)= 1$ means that $a$ and $b$ have no prime factors in common. $a^n$ has only the prime factors that $a$ has and $b^m$ has only the prime factors that $b$ has.



                        So $a^n$ and $b^n$ have no prime factors in common either. So $gcd(a^n, b^m) = 1$.



                        Simple as that.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          That seems very complicated and I'd give up.



                          If $gcd(a,b)= 1$ means that $a$ and $b$ have no prime factors in common. $a^n$ has only the prime factors that $a$ has and $b^m$ has only the prime factors that $b$ has.



                          So $a^n$ and $b^n$ have no prime factors in common either. So $gcd(a^n, b^m) = 1$.



                          Simple as that.






                          share|cite|improve this answer









                          $endgroup$



                          That seems very complicated and I'd give up.



                          If $gcd(a,b)= 1$ means that $a$ and $b$ have no prime factors in common. $a^n$ has only the prime factors that $a$ has and $b^m$ has only the prime factors that $b$ has.



                          So $a^n$ and $b^n$ have no prime factors in common either. So $gcd(a^n, b^m) = 1$.



                          Simple as that.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 6 '18 at 22:37









                          fleabloodfleablood

                          70.4k22685




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