Prove $gcd(a,m) mid gcd(ab,m)$ $forall a,b,m in Bbb Z$












1












$begingroup$


I named $gcd(a,m) = d$ and $gcd(ab,m) = d' $



So I know that $dmid a$, $dmid m $ and $d'mid ab $ , $d' mid m$



But I can't use the transitive property of divisibility here.



How can I prove that $d mid d'$?






elementary-number-theory divisibility greatest-common-divisor







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  • 1




    $begingroup$
    If $gmid a$ and $gmid m$, then $gmid ab$ and $gmid m$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 27 at 10:37
















1












$begingroup$


I named $gcd(a,m) = d$ and $gcd(ab,m) = d' $



So I know that $dmid a$, $dmid m $ and $d'mid ab $ , $d' mid m$



But I can't use the transitive property of divisibility here.



How can I prove that $d mid d'$?






elementary-number-theory divisibility greatest-common-divisor







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If $gmid a$ and $gmid m$, then $gmid ab$ and $gmid m$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 27 at 10:37














1












1








1





$begingroup$


I named $gcd(a,m) = d$ and $gcd(ab,m) = d' $



So I know that $dmid a$, $dmid m $ and $d'mid ab $ , $d' mid m$



But I can't use the transitive property of divisibility here.



How can I prove that $d mid d'$?






elementary-number-theory divisibility greatest-common-divisor







share|cite|improve this question











$endgroup$




I named $gcd(a,m) = d$ and $gcd(ab,m) = d' $



So I know that $dmid a$, $dmid m $ and $d'mid ab $ , $d' mid m$



But I can't use the transitive property of divisibility here.



How can I prove that $d mid d'$?






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 Jan 29 at 20:03









J. W. Tanner

3,7451320




3,7451320










asked Jan 27 at 10:25









Computer DreamComputer Dream

84




84








  • 1




    $begingroup$
    If $gmid a$ and $gmid m$, then $gmid ab$ and $gmid m$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 27 at 10:37














  • 1




    $begingroup$
    If $gmid a$ and $gmid m$, then $gmid ab$ and $gmid m$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 27 at 10:37








1




1




$begingroup$
If $gmid a$ and $gmid m$, then $gmid ab$ and $gmid m$.
$endgroup$
– Lord Shark the Unknown
Jan 27 at 10:37




$begingroup$
If $gmid a$ and $gmid m$, then $gmid ab$ and $gmid m$.
$endgroup$
– Lord Shark the Unknown
Jan 27 at 10:37










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$begingroup$

For each $xmid gcd(a,m)$, $xmid a,m$ so $xmid ab,m$ and $xmidgcd(ab,m)$.






share|cite|improve this answer











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    1












    $begingroup$

    For each $xmid gcd(a,m)$, $xmid a,m$ so $xmid ab,m$ and $xmidgcd(ab,m)$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      For each $xmid gcd(a,m)$, $xmid a,m$ so $xmid ab,m$ and $xmidgcd(ab,m)$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        For each $xmid gcd(a,m)$, $xmid a,m$ so $xmid ab,m$ and $xmidgcd(ab,m)$.






        share|cite|improve this answer











        $endgroup$



        For each $xmid gcd(a,m)$, $xmid a,m$ so $xmid ab,m$ and $xmidgcd(ab,m)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 27 at 10:44

























        answered Jan 27 at 10:30









        YiFanYiFan

        4,7861727




        4,7861727






























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