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Under which conditions is $gcd(a+bx,c)=1$ solvable and what is the solution?
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Let $a,b,cinmathbb{Z}$, $cneq0$. When is $gcd(a+bx,c)=1$ solvable and what is ${xinmathbb{Z}midgcd(a+bx,c)=1}$? A sufficient condition appears to be $gcd(a,b)=1$ but it is not necessary as $a=65$, $b=40$, $c=52$, and $x=1$ shows.
divisibility diophantine-equations greatest-common-divisor
asked Jan 25 at 13:26
MarioMario
562212
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add a comment |
$begingroup$
Let $a,b,cinmathbb{Z}$, $cneq0$. When is $gcd(a+bx,c)=1$ solvable and what is ${xinmathbb{Z}midgcd(a+bx,c)=1}$? A sufficient condition appears to be $gcd(a,b)=1$ but it is not necessary as $a=65$, $b=40$, $c=52$, and $x=1$ shows.
divisibility diophantine-equations greatest-common-divisor
asked Jan 25 at 13:26
MarioMario
562212
$endgroup$
add a comment |
$begingroup$
Let $a,b,cinmathbb{Z}$, $cneq0$. When is $gcd(a+bx,c)=1$ solvable and what is ${xinmathbb{Z}midgcd(a+bx,c)=1}$? A sufficient condition appears to be $gcd(a,b)=1$ but it is not necessary as $a=65$, $b=40$, $c=52$, and $x=1$ shows.
divisibility diophantine-equations greatest-common-divisor
asked Jan 25 at 13:26
MarioMario
562212
$endgroup$
Let $a,b,cinmathbb{Z}$, $cneq0$. When is $gcd(a+bx,c)=1$ solvable and what is ${xinmathbb{Z}midgcd(a+bx,c)=1}$? A sufficient condition appears to be $gcd(a,b)=1$ but it is not necessary as $a=65$, $b=40$, $c=52$, and $x=1$ shows.
divisibility diophantine-equations greatest-common-divisor
divisibility diophantine-equations greatest-common-divisor
asked Jan 25 at 13:26
MarioMario
562212
asked Jan 25 at 13:26
MarioMario
562212
asked Jan 25 at 13:26
MarioMario
562212
asked Jan 25 at 13:26
MarioMario
562212
asked Jan 25 at 13:26
MarioMario
562212
562212
add a comment |
add a comment |
1 Answer
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There exists $x$ with $,gcd(a+bx,c)=1iff gcd(a,b,c) = 1.,$ The direction $(Leftarrow)$ is in this answer, and $(Rightarrow)$ is clear, by $,gcd(a,b,c)mid a+bx,c,$ so also $,gcd(a+bx,c)$
answered Jan 25 at 14:35
Bill DubuqueBill Dubuque
212k29195654
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
There exists $x$ with $,gcd(a+bx,c)=1iff gcd(a,b,c) = 1.,$ The direction $(Leftarrow)$ is in this answer, and $(Rightarrow)$ is clear, by $,gcd(a,b,c)mid a+bx,c,$ so also $,gcd(a+bx,c)$
answered Jan 25 at 14:35
Bill DubuqueBill Dubuque
212k29195654
$endgroup$
add a comment |
$begingroup$
There exists $x$ with $,gcd(a+bx,c)=1iff gcd(a,b,c) = 1.,$ The direction $(Leftarrow)$ is in this answer, and $(Rightarrow)$ is clear, by $,gcd(a,b,c)mid a+bx,c,$ so also $,gcd(a+bx,c)$
answered Jan 25 at 14:35
Bill DubuqueBill Dubuque
212k29195654
$endgroup$
add a comment |
$begingroup$
There exists $x$ with $,gcd(a+bx,c)=1iff gcd(a,b,c) = 1.,$ The direction $(Leftarrow)$ is in this answer, and $(Rightarrow)$ is clear, by $,gcd(a,b,c)mid a+bx,c,$ so also $,gcd(a+bx,c)$
answered Jan 25 at 14:35
Bill DubuqueBill Dubuque
212k29195654
$endgroup$
There exists $x$ with $,gcd(a+bx,c)=1iff gcd(a,b,c) = 1.,$ The direction $(Leftarrow)$ is in this answer, and $(Rightarrow)$ is clear, by $,gcd(a,b,c)mid a+bx,c,$ so also $,gcd(a+bx,c)$
answered Jan 25 at 14:35
Bill DubuqueBill Dubuque
212k29195654
answered Jan 25 at 14:35
Bill DubuqueBill Dubuque
212k29195654
answered Jan 25 at 14:35
Bill DubuqueBill Dubuque
212k29195654
answered Jan 25 at 14:35
Bill DubuqueBill Dubuque
212k29195654
212k29195654
add a comment |
add a comment |
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