$(m,n)=1$, what could $(3n-4m, 5n+m)$ be?












3












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This is what I have so far:
let's say $d$ is the common divisor of $3n-4m$ and $5n+m$, then $d$ divides $5(3n-4m)-3(5n+m)=-23m$ and $3n-4m+4(5n+m)=23n$. $d$ can't be a divisor of both $m$ and $n$, so 23 must divide $d$. But is this the greatest common divisor of $3n-4m$ and $5n+m$?






number-theory







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  • $begingroup$
    In my understanding your conclusion does not look correct, because $d$ can divide both $m$ and $n$, and in that case it is equal to $1$. Otherwise, it divides $23$, hence it is either $1$ or $23$.
    $endgroup$
    – Gibbs
    Dec 6 '18 at 23:41










  • $begingroup$
    What does $(m/n)$ mean???
    $endgroup$
    – bof
    Dec 6 '18 at 23:42










  • $begingroup$
    I meant the greatest common divisor.
    $endgroup$
    – Lowkey
    Dec 6 '18 at 23:45










  • $begingroup$
    Um don't you mean $d$ must divide $23$. Not $23$ must divide $d$?
    $endgroup$
    – fleablood
    Dec 7 '18 at 1:33
















3












$begingroup$


This is what I have so far:
let's say $d$ is the common divisor of $3n-4m$ and $5n+m$, then $d$ divides $5(3n-4m)-3(5n+m)=-23m$ and $3n-4m+4(5n+m)=23n$. $d$ can't be a divisor of both $m$ and $n$, so 23 must divide $d$. But is this the greatest common divisor of $3n-4m$ and $5n+m$?






number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    In my understanding your conclusion does not look correct, because $d$ can divide both $m$ and $n$, and in that case it is equal to $1$. Otherwise, it divides $23$, hence it is either $1$ or $23$.
    $endgroup$
    – Gibbs
    Dec 6 '18 at 23:41










  • $begingroup$
    What does $(m/n)$ mean???
    $endgroup$
    – bof
    Dec 6 '18 at 23:42










  • $begingroup$
    I meant the greatest common divisor.
    $endgroup$
    – Lowkey
    Dec 6 '18 at 23:45










  • $begingroup$
    Um don't you mean $d$ must divide $23$. Not $23$ must divide $d$?
    $endgroup$
    – fleablood
    Dec 7 '18 at 1:33














3












3








3





$begingroup$


This is what I have so far:
let's say $d$ is the common divisor of $3n-4m$ and $5n+m$, then $d$ divides $5(3n-4m)-3(5n+m)=-23m$ and $3n-4m+4(5n+m)=23n$. $d$ can't be a divisor of both $m$ and $n$, so 23 must divide $d$. But is this the greatest common divisor of $3n-4m$ and $5n+m$?






number-theory







share|cite|improve this question











$endgroup$




This is what I have so far:
let's say $d$ is the common divisor of $3n-4m$ and $5n+m$, then $d$ divides $5(3n-4m)-3(5n+m)=-23m$ and $3n-4m+4(5n+m)=23n$. $d$ can't be a divisor of both $m$ and $n$, so 23 must divide $d$. But is this the greatest common divisor of $3n-4m$ and $5n+m$?






number-theory




number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 23:43







Lowkey

















asked Dec 6 '18 at 23:25









LowkeyLowkey

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728












  • $begingroup$
    In my understanding your conclusion does not look correct, because $d$ can divide both $m$ and $n$, and in that case it is equal to $1$. Otherwise, it divides $23$, hence it is either $1$ or $23$.
    $endgroup$
    – Gibbs
    Dec 6 '18 at 23:41










  • $begingroup$
    What does $(m/n)$ mean???
    $endgroup$
    – bof
    Dec 6 '18 at 23:42










  • $begingroup$
    I meant the greatest common divisor.
    $endgroup$
    – Lowkey
    Dec 6 '18 at 23:45










  • $begingroup$
    Um don't you mean $d$ must divide $23$. Not $23$ must divide $d$?
    $endgroup$
    – fleablood
    Dec 7 '18 at 1:33


















  • $begingroup$
    In my understanding your conclusion does not look correct, because $d$ can divide both $m$ and $n$, and in that case it is equal to $1$. Otherwise, it divides $23$, hence it is either $1$ or $23$.
    $endgroup$
    – Gibbs
    Dec 6 '18 at 23:41










  • $begingroup$
    What does $(m/n)$ mean???
    $endgroup$
    – bof
    Dec 6 '18 at 23:42










  • $begingroup$
    I meant the greatest common divisor.
    $endgroup$
    – Lowkey
    Dec 6 '18 at 23:45










  • $begingroup$
    Um don't you mean $d$ must divide $23$. Not $23$ must divide $d$?
    $endgroup$
    – fleablood
    Dec 7 '18 at 1:33
















$begingroup$
In my understanding your conclusion does not look correct, because $d$ can divide both $m$ and $n$, and in that case it is equal to $1$. Otherwise, it divides $23$, hence it is either $1$ or $23$.
$endgroup$
– Gibbs
Dec 6 '18 at 23:41




$begingroup$
In my understanding your conclusion does not look correct, because $d$ can divide both $m$ and $n$, and in that case it is equal to $1$. Otherwise, it divides $23$, hence it is either $1$ or $23$.
$endgroup$
– Gibbs
Dec 6 '18 at 23:41












$begingroup$
What does $(m/n)$ mean???
$endgroup$
– bof
Dec 6 '18 at 23:42




$begingroup$
What does $(m/n)$ mean???
$endgroup$
– bof
Dec 6 '18 at 23:42












$begingroup$
I meant the greatest common divisor.
$endgroup$
– Lowkey
Dec 6 '18 at 23:45




$begingroup$
I meant the greatest common divisor.
$endgroup$
– Lowkey
Dec 6 '18 at 23:45












$begingroup$
Um don't you mean $d$ must divide $23$. Not $23$ must divide $d$?
$endgroup$
– fleablood
Dec 7 '18 at 1:33




$begingroup$
Um don't you mean $d$ must divide $23$. Not $23$ must divide $d$?
$endgroup$
– fleablood
Dec 7 '18 at 1:33










3 Answers
3






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oldest

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1












$begingroup$

You have the right idea - eliminate $n$ and $m,,$ but your conclusion is misworded (reversed). It should state that $23$ must be divisible by $d$ (not $23$ must divide $d$) since $$,dmid 23m,23niff dmid (23m,23n)! =! 23(m,n)! =! 23 {rm by} (m,n)=1$$



Remark $ $ We can use linear algebra language, i.e. use Cramer's rule to solve for $,m,n,$ below



$qquadqquadqquad$ $ begin{eqnarray}
3 n, - 4,m & = & i\[.3em]
5,n +, 1,m &=& j end{eqnarray}quadRightarrowquad
$ $
begin{align}
23,n &= , i +, 4,j \[.3em]
23,m, &=, -5,i + 3,j end{align} $



Thus by RHS: $,dmid i,j,Rightarrow, dmid 23m,23n,Rightarrow, dmid 23 $ as above. More generally this method yields



Theorem $ $ If $rm,[x,y]overset{A}mapsto [X,Y],$ is linear then $: rm(x,y)mid (X,Y)mid color{#90f}Delta, (x,y), color{#90f}{Delta := {rm det}, A}$



e.g. $ $ in OP we have $,color{#90f}{Delta =bf 23},$ so the above yields $ (3n!-!4m,5n!+!m)midcolor{#90f}{bf 23}(n,m) = 23$

using the map $ [n,m]mapsto [3n!-!4m,,5n!+!m] $ i.e.



$$ [n,,m],mapsto, [n,,m],underbrace{begin{bmatrix}3 & 5\ -4 &1end{bmatrix}}_{large det A, =, color{#90f}{bf 23}} =, [3n!-!4m,,5n!+!m]$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for explaining where I went wrong, and then providing more interesting information!
    $endgroup$
    – Lowkey
    Dec 7 '18 at 1:46



















3












$begingroup$

If $m=3$ and $n=4$, the greatest common divisor of $3n-4m=0$ and $5n+m=23$ is $23$.



If $m=1$ and $n=1$, the greatest common divisor of $3n-4m=-1$ and $5n+m=6$ is $1$.



Your argument is good, but the conclusion is wrong: you prove that $d$ can be a divisor of $23$. Thus it is either $1$ or $23$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, this clears things up.
    $endgroup$
    – Lowkey
    Dec 6 '18 at 23:49



















2












$begingroup$

By Bezout we have $An+Bm=1$, multiply this by $23$ and create some terms
begin{eqnarray*}
20An+4Am-4Am+3An+20Bm-15Bn+15Bn+3Bm=23.
end{eqnarray*}
Now factorise
begin{eqnarray*}
(4A+3B)(m+5n) +(A-5B)(3n-4m) =23.
end{eqnarray*}
So if $(4A+3B,A-5B)=1$ then $(m+5n,3n-4m)=23$ and



and if $(4A+3B,A-5B)=23$ then $(m+5n,3n-4m)=1$.






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    3 Answers
    3






    active

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    3 Answers
    3






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    You have the right idea - eliminate $n$ and $m,,$ but your conclusion is misworded (reversed). It should state that $23$ must be divisible by $d$ (not $23$ must divide $d$) since $$,dmid 23m,23niff dmid (23m,23n)! =! 23(m,n)! =! 23 {rm by} (m,n)=1$$



    Remark $ $ We can use linear algebra language, i.e. use Cramer's rule to solve for $,m,n,$ below



    $qquadqquadqquad$ $ begin{eqnarray}
    3 n, - 4,m & = & i\[.3em]
    5,n +, 1,m &=& j end{eqnarray}quadRightarrowquad
    $     $
    begin{align}
    23,n &= , i +, 4,j \[.3em]
    23,m, &=, -5,i + 3,j end{align} $



    Thus by RHS: $,dmid i,j,Rightarrow, dmid 23m,23n,Rightarrow, dmid 23 $ as above. More generally this method yields



    Theorem $ $ If $rm,[x,y]overset{A}mapsto [X,Y],$ is linear then $: rm(x,y)mid (X,Y)mid color{#90f}Delta, (x,y), color{#90f}{Delta := {rm det}, A}$



    e.g. $ $ in OP we have $,color{#90f}{Delta =bf 23},$ so the above yields $ (3n!-!4m,5n!+!m)midcolor{#90f}{bf 23}(n,m) = 23$

    using the map $ [n,m]mapsto [3n!-!4m,,5n!+!m] $ i.e.



    $$ [n,,m],mapsto, [n,,m],underbrace{begin{bmatrix}3 & 5\ -4 &1end{bmatrix}}_{large det A, =, color{#90f}{bf 23}} =, [3n!-!4m,,5n!+!m]$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for explaining where I went wrong, and then providing more interesting information!
      $endgroup$
      – Lowkey
      Dec 7 '18 at 1:46
















    1












    $begingroup$

    You have the right idea - eliminate $n$ and $m,,$ but your conclusion is misworded (reversed). It should state that $23$ must be divisible by $d$ (not $23$ must divide $d$) since $$,dmid 23m,23niff dmid (23m,23n)! =! 23(m,n)! =! 23 {rm by} (m,n)=1$$



    Remark $ $ We can use linear algebra language, i.e. use Cramer's rule to solve for $,m,n,$ below



    $qquadqquadqquad$ $ begin{eqnarray}
    3 n, - 4,m & = & i\[.3em]
    5,n +, 1,m &=& j end{eqnarray}quadRightarrowquad
    $     $
    begin{align}
    23,n &= , i +, 4,j \[.3em]
    23,m, &=, -5,i + 3,j end{align} $



    Thus by RHS: $,dmid i,j,Rightarrow, dmid 23m,23n,Rightarrow, dmid 23 $ as above. More generally this method yields



    Theorem $ $ If $rm,[x,y]overset{A}mapsto [X,Y],$ is linear then $: rm(x,y)mid (X,Y)mid color{#90f}Delta, (x,y), color{#90f}{Delta := {rm det}, A}$



    e.g. $ $ in OP we have $,color{#90f}{Delta =bf 23},$ so the above yields $ (3n!-!4m,5n!+!m)midcolor{#90f}{bf 23}(n,m) = 23$

    using the map $ [n,m]mapsto [3n!-!4m,,5n!+!m] $ i.e.



    $$ [n,,m],mapsto, [n,,m],underbrace{begin{bmatrix}3 & 5\ -4 &1end{bmatrix}}_{large det A, =, color{#90f}{bf 23}} =, [3n!-!4m,,5n!+!m]$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for explaining where I went wrong, and then providing more interesting information!
      $endgroup$
      – Lowkey
      Dec 7 '18 at 1:46














    1












    1








    1





    $begingroup$

    You have the right idea - eliminate $n$ and $m,,$ but your conclusion is misworded (reversed). It should state that $23$ must be divisible by $d$ (not $23$ must divide $d$) since $$,dmid 23m,23niff dmid (23m,23n)! =! 23(m,n)! =! 23 {rm by} (m,n)=1$$



    Remark $ $ We can use linear algebra language, i.e. use Cramer's rule to solve for $,m,n,$ below



    $qquadqquadqquad$ $ begin{eqnarray}
    3 n, - 4,m & = & i\[.3em]
    5,n +, 1,m &=& j end{eqnarray}quadRightarrowquad
    $     $
    begin{align}
    23,n &= , i +, 4,j \[.3em]
    23,m, &=, -5,i + 3,j end{align} $



    Thus by RHS: $,dmid i,j,Rightarrow, dmid 23m,23n,Rightarrow, dmid 23 $ as above. More generally this method yields



    Theorem $ $ If $rm,[x,y]overset{A}mapsto [X,Y],$ is linear then $: rm(x,y)mid (X,Y)mid color{#90f}Delta, (x,y), color{#90f}{Delta := {rm det}, A}$



    e.g. $ $ in OP we have $,color{#90f}{Delta =bf 23},$ so the above yields $ (3n!-!4m,5n!+!m)midcolor{#90f}{bf 23}(n,m) = 23$

    using the map $ [n,m]mapsto [3n!-!4m,,5n!+!m] $ i.e.



    $$ [n,,m],mapsto, [n,,m],underbrace{begin{bmatrix}3 & 5\ -4 &1end{bmatrix}}_{large det A, =, color{#90f}{bf 23}} =, [3n!-!4m,,5n!+!m]$$






    share|cite|improve this answer











    $endgroup$



    You have the right idea - eliminate $n$ and $m,,$ but your conclusion is misworded (reversed). It should state that $23$ must be divisible by $d$ (not $23$ must divide $d$) since $$,dmid 23m,23niff dmid (23m,23n)! =! 23(m,n)! =! 23 {rm by} (m,n)=1$$



    Remark $ $ We can use linear algebra language, i.e. use Cramer's rule to solve for $,m,n,$ below



    $qquadqquadqquad$ $ begin{eqnarray}
    3 n, - 4,m & = & i\[.3em]
    5,n +, 1,m &=& j end{eqnarray}quadRightarrowquad
    $     $
    begin{align}
    23,n &= , i +, 4,j \[.3em]
    23,m, &=, -5,i + 3,j end{align} $



    Thus by RHS: $,dmid i,j,Rightarrow, dmid 23m,23n,Rightarrow, dmid 23 $ as above. More generally this method yields



    Theorem $ $ If $rm,[x,y]overset{A}mapsto [X,Y],$ is linear then $: rm(x,y)mid (X,Y)mid color{#90f}Delta, (x,y), color{#90f}{Delta := {rm det}, A}$



    e.g. $ $ in OP we have $,color{#90f}{Delta =bf 23},$ so the above yields $ (3n!-!4m,5n!+!m)midcolor{#90f}{bf 23}(n,m) = 23$

    using the map $ [n,m]mapsto [3n!-!4m,,5n!+!m] $ i.e.



    $$ [n,,m],mapsto, [n,,m],underbrace{begin{bmatrix}3 & 5\ -4 &1end{bmatrix}}_{large det A, =, color{#90f}{bf 23}} =, [3n!-!4m,,5n!+!m]$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 7 '18 at 20:38

























    answered Dec 7 '18 at 1:19









    Bill DubuqueBill Dubuque

    209k29190633




    209k29190633












    • $begingroup$
      Thanks for explaining where I went wrong, and then providing more interesting information!
      $endgroup$
      – Lowkey
      Dec 7 '18 at 1:46


















    • $begingroup$
      Thanks for explaining where I went wrong, and then providing more interesting information!
      $endgroup$
      – Lowkey
      Dec 7 '18 at 1:46
















    $begingroup$
    Thanks for explaining where I went wrong, and then providing more interesting information!
    $endgroup$
    – Lowkey
    Dec 7 '18 at 1:46




    $begingroup$
    Thanks for explaining where I went wrong, and then providing more interesting information!
    $endgroup$
    – Lowkey
    Dec 7 '18 at 1:46











    3












    $begingroup$

    If $m=3$ and $n=4$, the greatest common divisor of $3n-4m=0$ and $5n+m=23$ is $23$.



    If $m=1$ and $n=1$, the greatest common divisor of $3n-4m=-1$ and $5n+m=6$ is $1$.



    Your argument is good, but the conclusion is wrong: you prove that $d$ can be a divisor of $23$. Thus it is either $1$ or $23$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks, this clears things up.
      $endgroup$
      – Lowkey
      Dec 6 '18 at 23:49
















    3












    $begingroup$

    If $m=3$ and $n=4$, the greatest common divisor of $3n-4m=0$ and $5n+m=23$ is $23$.



    If $m=1$ and $n=1$, the greatest common divisor of $3n-4m=-1$ and $5n+m=6$ is $1$.



    Your argument is good, but the conclusion is wrong: you prove that $d$ can be a divisor of $23$. Thus it is either $1$ or $23$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks, this clears things up.
      $endgroup$
      – Lowkey
      Dec 6 '18 at 23:49














    3












    3








    3





    $begingroup$

    If $m=3$ and $n=4$, the greatest common divisor of $3n-4m=0$ and $5n+m=23$ is $23$.



    If $m=1$ and $n=1$, the greatest common divisor of $3n-4m=-1$ and $5n+m=6$ is $1$.



    Your argument is good, but the conclusion is wrong: you prove that $d$ can be a divisor of $23$. Thus it is either $1$ or $23$.






    share|cite|improve this answer









    $endgroup$



    If $m=3$ and $n=4$, the greatest common divisor of $3n-4m=0$ and $5n+m=23$ is $23$.



    If $m=1$ and $n=1$, the greatest common divisor of $3n-4m=-1$ and $5n+m=6$ is $1$.



    Your argument is good, but the conclusion is wrong: you prove that $d$ can be a divisor of $23$. Thus it is either $1$ or $23$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 6 '18 at 23:42









    egregegreg

    180k1485202




    180k1485202












    • $begingroup$
      Thanks, this clears things up.
      $endgroup$
      – Lowkey
      Dec 6 '18 at 23:49


















    • $begingroup$
      Thanks, this clears things up.
      $endgroup$
      – Lowkey
      Dec 6 '18 at 23:49
















    $begingroup$
    Thanks, this clears things up.
    $endgroup$
    – Lowkey
    Dec 6 '18 at 23:49




    $begingroup$
    Thanks, this clears things up.
    $endgroup$
    – Lowkey
    Dec 6 '18 at 23:49











    2












    $begingroup$

    By Bezout we have $An+Bm=1$, multiply this by $23$ and create some terms
    begin{eqnarray*}
    20An+4Am-4Am+3An+20Bm-15Bn+15Bn+3Bm=23.
    end{eqnarray*}
    Now factorise
    begin{eqnarray*}
    (4A+3B)(m+5n) +(A-5B)(3n-4m) =23.
    end{eqnarray*}
    So if $(4A+3B,A-5B)=1$ then $(m+5n,3n-4m)=23$ and



    and if $(4A+3B,A-5B)=23$ then $(m+5n,3n-4m)=1$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      By Bezout we have $An+Bm=1$, multiply this by $23$ and create some terms
      begin{eqnarray*}
      20An+4Am-4Am+3An+20Bm-15Bn+15Bn+3Bm=23.
      end{eqnarray*}
      Now factorise
      begin{eqnarray*}
      (4A+3B)(m+5n) +(A-5B)(3n-4m) =23.
      end{eqnarray*}
      So if $(4A+3B,A-5B)=1$ then $(m+5n,3n-4m)=23$ and



      and if $(4A+3B,A-5B)=23$ then $(m+5n,3n-4m)=1$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        By Bezout we have $An+Bm=1$, multiply this by $23$ and create some terms
        begin{eqnarray*}
        20An+4Am-4Am+3An+20Bm-15Bn+15Bn+3Bm=23.
        end{eqnarray*}
        Now factorise
        begin{eqnarray*}
        (4A+3B)(m+5n) +(A-5B)(3n-4m) =23.
        end{eqnarray*}
        So if $(4A+3B,A-5B)=1$ then $(m+5n,3n-4m)=23$ and



        and if $(4A+3B,A-5B)=23$ then $(m+5n,3n-4m)=1$.






        share|cite|improve this answer









        $endgroup$



        By Bezout we have $An+Bm=1$, multiply this by $23$ and create some terms
        begin{eqnarray*}
        20An+4Am-4Am+3An+20Bm-15Bn+15Bn+3Bm=23.
        end{eqnarray*}
        Now factorise
        begin{eqnarray*}
        (4A+3B)(m+5n) +(A-5B)(3n-4m) =23.
        end{eqnarray*}
        So if $(4A+3B,A-5B)=1$ then $(m+5n,3n-4m)=23$ and



        and if $(4A+3B,A-5B)=23$ then $(m+5n,3n-4m)=1$.







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        answered Dec 6 '18 at 23:47









        Donald SplutterwitDonald Splutterwit

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