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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0There's a theorem which states that\[\gcd(a,b)=\gcd(a+kb,b)=\gcd(a,b+qa)\]where \(a,b,k,q\in\mathbb{Z}\).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So, obviously,\[\gcd(a,b)=\gcd(a+b,b)=\gcd(a,ab).\]But I get stuck there. :/

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0When 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), (ab)

watchmath
 4 years ago
Best ResponseYou've already chosen the best response.2(roughly ) if p divides a+b and ab then p divides 2a and also 2b. So either p is 2 or p divides a and b, which implies p=1.

watchmath
 4 years ago
Best ResponseYou've already chosen the best response.2maybe 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0james, this is what i did: gcd(a,b)=1 pa+qb=1 pa+pb+qa2qb+qb=1+pb+qa2qb p(a+b)+q(ab)=1+[pb+q(a2b)] however, what i surrounded in parenthesis on the rhs is gcd(b,a2b). but using the theorem i specified in my first post, we know that gcd(b,a2b)=1. so p(a+b)+q(ab)=1+1=2 but how can i show that it could also be 1? :/

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0because the choice of p,q for the relation (a,b)=1 need not be the same p,q for the relation (a+b,ab)=1. watch math's proof is perfectly good.
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