Question

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?\[\]

Answer 1

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

Answer 2

So, obviously,\[\gcd(a,b)=\gcd(a+b,b)=\gcd(a,a-b).\]But I get stuck there. :/

Answer 3

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)

Answer 4

(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.

Answer 5

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.

Answer 6

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? :/

Answer 7

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.