## 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?

1. anonymous

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

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

3. JamesJ

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

(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

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

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

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.