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anonymous

  • 4 years ago

Let \(a,b\in\mathbb{Z}\). Then show that \(\gcd(a,b)=1\implies\gcd(a+b,a-b)=1\) or \(2\). How could I do this?\[\]

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  1. anonymous
    • 4 years ago
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    There's a theorem which states that\[\gcd(a,b)=\gcd(a+kb,b)=\gcd(a,b+qa)\]where \(a,b,k,q\in\mathbb{Z}\).

  2. anonymous
    • 4 years ago
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    So, obviously,\[\gcd(a,b)=\gcd(a+b,b)=\gcd(a,a-b).\]But I get stuck there. :/

  3. JamesJ
    • 4 years ago
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    When you're stuck, go back to basics. (a,b) = 1, if and only if there exist integers p, q such that pa + qb = 1 Now see if you can manipulate that to get an expression in (a+b), (a-b)

  4. watchmath
    • 4 years ago
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    (roughly ) if p divides a+b and a-b then p divides 2a and also 2b. So either p is 2 or p divides a and b, which implies p=1.

  5. watchmath
    • 4 years ago
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    maybe to be more accurate we should say like this. Suppose the gcd is not 1, then it has a prime factor p. But then it implies p divides 2a and p divides 2b. If p not divide 2, then p divides a and b and hence p divides 1 (contradiction1). So p divides 2, which implies p=2.

  6. anonymous
    • 4 years ago
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    james, this is what i did: gcd(a,b)=1 pa+qb=1 pa+pb+qa-2qb+qb=1+pb+qa-2qb p(a+b)+q(a-b)=1+[pb+q(a-2b)] however, what i surrounded in parenthesis on the rhs is gcd(b,a-2b). but using the theorem i specified in my first post, we know that gcd(b,a-2b)=1. so p(a+b)+q(a-b)=1+1=2 but how can i show that it could also be 1? :/

  7. JamesJ
    • 4 years ago
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    because the choice of p,q for the relation (a,b)=1 need not be the same p,q for the relation (a+b,a-b)=1. watch math's proof is perfectly good.

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