$(m,n)=1$, what could $(3n-4m, 5n+m)$ be?
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
asked Dec 6 at 23:25
Lowkey
728
add a comment |
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
asked Dec 6 at 23:25
Lowkey
728
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 $23$.
– Gibbs
Dec 6 at 23:41
What does $(m/n)$ mean???
– bof
Dec 6 at 23:42
I meant the greatest common divisor.
– Lowkey
Dec 6 at 23:45
Um don't you mean $d$ must divide $23$. Not $23$ must divide $d$?
– fleablood
Dec 7 at 1:33
add a comment |
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
asked Dec 6 at 23:25
Lowkey
728
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
asked Dec 6 at 23:25
Lowkey
728
asked Dec 6 at 23:25
Lowkey
728
edited Dec 6 at 23:43
asked Dec 6 at 23:25
Lowkey
728
asked Dec 6 at 23:25
Lowkey
728
asked Dec 6 at 23:25
Lowkey
728
728
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 $23$.
– Gibbs
Dec 6 at 23:41
What does $(m/n)$ mean???
– bof
Dec 6 at 23:42
I meant the greatest common divisor.
– Lowkey
Dec 6 at 23:45
Um don't you mean $d$ must divide $23$. Not $23$ must divide $d$?
– fleablood
Dec 7 at 1:33
add a comment |
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 $23$.
– Gibbs
Dec 6 at 23:41
What does $(m/n)$ mean???
– bof
Dec 6 at 23:42
I meant the greatest common divisor.
– Lowkey
Dec 6 at 23:45
Um don't you mean $d$ must divide $23$. Not $23$ must divide $d$?
– fleablood
Dec 7 at 1:33
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 $23$.
– Gibbs
Dec 6 at 23:41
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 $23$.
– Gibbs
Dec 6 at 23:41
What does $(m/n)$ mean???
– bof
Dec 6 at 23:42
What does $(m/n)$ mean???
– bof
Dec 6 at 23:42
I meant the greatest common divisor.
– Lowkey
Dec 6 at 23:45
I meant the greatest common divisor.
– Lowkey
Dec 6 at 23:45
Um don't you mean $d$ must divide $23$. Not $23$ must divide $d$?
– fleablood
Dec 7 at 1:33
Um don't you mean $d$ must divide $23$. Not $23$ must divide $d$?
– fleablood
Dec 7 at 1:33
add a comment |
3 Answers
3
active
oldest
votes
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.
$$ [n,,m],mapsto, [n,,m],underbrace{begin{bmatrix}3 & 5\ -4 &1end{bmatrix}}_{large det A, =, color{#90f}{bf 23}} =, [3n!-!4m,,5n!+!m]$$
answered Dec 7 at 1:19
Bill Dubuque
208k29190626
Thanks for explaining where I went wrong, and then providing more interesting information!
– Lowkey
Dec 7 at 1:46
add a comment |
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 $23$.
answered Dec 6 at 23:42
egreg
177k1484198
Thanks, this clears things up.
– Lowkey
Dec 6 at 23:49
add a comment |
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 $(m+5n,3n-4m)=1$.
answered Dec 6 at 23:47
Donald Splutterwit
22.3k21244
add a comment |
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3 Answers
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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.
$$ [n,,m],mapsto, [n,,m],underbrace{begin{bmatrix}3 & 5\ -4 &1end{bmatrix}}_{large det A, =, color{#90f}{bf 23}} =, [3n!-!4m,,5n!+!m]$$
answered Dec 7 at 1:19
Bill Dubuque
208k29190626
Thanks for explaining where I went wrong, and then providing more interesting information!
– Lowkey
Dec 7 at 1:46
add a comment |
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.
$$ [n,,m],mapsto, [n,,m],underbrace{begin{bmatrix}3 & 5\ -4 &1end{bmatrix}}_{large det A, =, color{#90f}{bf 23}} =, [3n!-!4m,,5n!+!m]$$
answered Dec 7 at 1:19
Bill Dubuque
208k29190626
Thanks for explaining where I went wrong, and then providing more interesting information!
– Lowkey
Dec 7 at 1:46
add a comment |
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.
$$ [n,,m],mapsto, [n,,m],underbrace{begin{bmatrix}3 & 5\ -4 &1end{bmatrix}}_{large det A, =, color{#90f}{bf 23}} =, [3n!-!4m,,5n!+!m]$$
answered Dec 7 at 1:19
Bill Dubuque
208k29190626
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.
$$ [n,,m],mapsto, [n,,m],underbrace{begin{bmatrix}3 & 5\ -4 &1end{bmatrix}}_{large det A, =, color{#90f}{bf 23}} =, [3n!-!4m,,5n!+!m]$$
answered Dec 7 at 1:19
Bill Dubuque
208k29190626
edited Dec 7 at 20:38
answered Dec 7 at 1:19
Bill Dubuque
208k29190626
answered Dec 7 at 1:19
Bill Dubuque
208k29190626
answered Dec 7 at 1:19
Bill Dubuque
208k29190626
208k29190626
Thanks for explaining where I went wrong, and then providing more interesting information!
– Lowkey
Dec 7 at 1:46
add a comment |
Thanks for explaining where I went wrong, and then providing more interesting information!
– Lowkey
Dec 7 at 1:46
Thanks for explaining where I went wrong, and then providing more interesting information!
– Lowkey
Dec 7 at 1:46
Thanks for explaining where I went wrong, and then providing more interesting information!
– Lowkey
Dec 7 at 1:46
add a comment |
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 $23$.
answered Dec 6 at 23:42
egreg
177k1484198
Thanks, this clears things up.
– Lowkey
Dec 6 at 23:49
add a comment |
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 $23$.
answered Dec 6 at 23:42
egreg
177k1484198
Thanks, this clears things up.
– Lowkey
Dec 6 at 23:49
add a comment |
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 $23$.
answered Dec 6 at 23:42
egreg
177k1484198
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 $23$.
answered Dec 6 at 23:42
egreg
177k1484198
answered Dec 6 at 23:42
egreg
177k1484198
answered Dec 6 at 23:42
egreg
177k1484198
answered Dec 6 at 23:42
egreg
177k1484198
177k1484198
Thanks, this clears things up.
– Lowkey
Dec 6 at 23:49
add a comment |
Thanks, this clears things up.
– Lowkey
Dec 6 at 23:49
Thanks, this clears things up.
– Lowkey
Dec 6 at 23:49
Thanks, this clears things up.
– Lowkey
Dec 6 at 23:49
add a comment |
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 $(m+5n,3n-4m)=1$.
answered Dec 6 at 23:47
Donald Splutterwit
22.3k21244
add a comment |
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 $(m+5n,3n-4m)=1$.
answered Dec 6 at 23:47
Donald Splutterwit
22.3k21244
add a comment |
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 $(m+5n,3n-4m)=1$.
answered Dec 6 at 23:47
Donald Splutterwit
22.3k21244
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 $(m+5n,3n-4m)=1$.
answered Dec 6 at 23:47
Donald Splutterwit
22.3k21244
answered Dec 6 at 23:47
Donald Splutterwit
22.3k21244
answered Dec 6 at 23:47
Donald Splutterwit
22.3k21244
answered Dec 6 at 23:47
Donald Splutterwit
22.3k21244
22.3k21244
add a comment |
add a comment |
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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 $23$.
– Gibbs
Dec 6 at 23:41
What does $(m/n)$ mean???
– bof
Dec 6 at 23:42
I meant the greatest common divisor.
– Lowkey
Dec 6 at 23:45
Um don't you mean $d$ must divide $23$. Not $23$ must divide $d$?
– fleablood
Dec 7 at 1:33