prove that if gcd(a,b) = 1, then gcd(a,a+b) = 1 [duplicate]
This question already has an answer here:
If $gcd(a,b) = 1$ then $gcd(a,a + b) = 1$
3 answers
So I need to prove that if gcd(a,b) = 1, then gcd(a,a+b) = 1. How do I go about it? I tried going with the ax + by = 1 method but it didn't work out so well.
elementary-number-theory
asked Nov 27 '18 at 1:14
geetarbui
1
marked as duplicate by Bill Dubuque elementary-number-theory
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Nov 27 '18 at 1:21
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This question already has an answer here:
If $gcd(a,b) = 1$ then $gcd(a,a + b) = 1$
3 answers
So I need to prove that if gcd(a,b) = 1, then gcd(a,a+b) = 1. How do I go about it? I tried going with the ax + by = 1 method but it didn't work out so well.
elementary-number-theory
asked Nov 27 '18 at 1:14
geetarbui
1
marked as duplicate by Bill Dubuque elementary-number-theory
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Nov 27 '18 at 1:21
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This question already has an answer here:
If $gcd(a,b) = 1$ then $gcd(a,a + b) = 1$
3 answers
So I need to prove that if gcd(a,b) = 1, then gcd(a,a+b) = 1. How do I go about it? I tried going with the ax + by = 1 method but it didn't work out so well.
elementary-number-theory
asked Nov 27 '18 at 1:14
geetarbui
1
This question already has an answer here:
If $gcd(a,b) = 1$ then $gcd(a,a + b) = 1$
3 answers
So I need to prove that if gcd(a,b) = 1, then gcd(a,a+b) = 1. How do I go about it? I tried going with the ax + by = 1 method but it didn't work out so well.
This question already has an answer here:
If $gcd(a,b) = 1$ then $gcd(a,a + b) = 1$
3 answers
elementary-number-theory
elementary-number-theory
asked Nov 27 '18 at 1:14
geetarbui
1
asked Nov 27 '18 at 1:14
geetarbui
1
asked Nov 27 '18 at 1:14
geetarbui
1
asked Nov 27 '18 at 1:14
geetarbui
1
asked Nov 27 '18 at 1:14
geetarbui
1
1
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Nov 27 '18 at 1:21
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1 Answer
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I think it should have worked out well, as
$$ax+by = 1$$
can be rewritten as
$$a(x-y) + (a+b)y = 1$$
And since $x-y$ is an integer, we can just call it $x'$, and then we have
$$ax'+(a+b)y = 1$$
answered Nov 27 '18 at 1:18
Isaac Browne
4,60731132
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1 Answer
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1 Answer
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I think it should have worked out well, as
$$ax+by = 1$$
can be rewritten as
$$a(x-y) + (a+b)y = 1$$
And since $x-y$ is an integer, we can just call it $x'$, and then we have
$$ax'+(a+b)y = 1$$
answered Nov 27 '18 at 1:18
Isaac Browne
4,60731132
add a comment |
I think it should have worked out well, as
$$ax+by = 1$$
can be rewritten as
$$a(x-y) + (a+b)y = 1$$
And since $x-y$ is an integer, we can just call it $x'$, and then we have
$$ax'+(a+b)y = 1$$
answered Nov 27 '18 at 1:18
Isaac Browne
4,60731132
add a comment |
I think it should have worked out well, as
$$ax+by = 1$$
can be rewritten as
$$a(x-y) + (a+b)y = 1$$
And since $x-y$ is an integer, we can just call it $x'$, and then we have
$$ax'+(a+b)y = 1$$
answered Nov 27 '18 at 1:18
Isaac Browne
4,60731132
I think it should have worked out well, as
$$ax+by = 1$$
can be rewritten as
$$a(x-y) + (a+b)y = 1$$
And since $x-y$ is an integer, we can just call it $x'$, and then we have
$$ax'+(a+b)y = 1$$
answered Nov 27 '18 at 1:18
Isaac Browne
4,60731132
answered Nov 27 '18 at 1:18
Isaac Browne
4,60731132
answered Nov 27 '18 at 1:18
Isaac Browne
4,60731132
answered Nov 27 '18 at 1:18
Isaac Browne
4,60731132
4,60731132
add a comment |
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