Why Does gcd(a+b, a-b) Only Equal 1 or 2?

[PsychonautQQ]

Homework Statement

Show that gcd(a+b,a-b) is either 1 or 2. (hint, show that d|2a and d|2b)

Homework Equations

d = x(a+b)+y(a-b)

The Attempt at a Solution

so by the definition of divisibility:
a+b = dr
a-b = ds

adding and subtracting these equalities from each other we can arrive at where the hint wanted us to conclude:
(r+s) = 2a/d
(r-s) = 2b/d

Trying to figure out what to do from here, having a hard time using the hint to restrict d to 1 or 2.

 

[Dick]

Homework Statement

Show that gcd(a+b,a-b) is either 1 or 2. (hint, show that d|2a and d|2b)

Homework Equations

d = x(a+b)+y(a-b)

The Attempt at a Solution

so by the definition of divisibility:
a+b = dr
a-b = ds

adding and subtracting these equalities from each other we can arrive at where the hint wanted us to conclude:
(r+s) = 2a/d
(r-s) = 2b/d

Trying to figure out what to do from here, having a hard time using the hint to restrict d to 1 or 2.

You must have some condition on a and b. Like, are they relatively prime?

 

[Rellek]

Hey there,

I think a really important concept that you might be forgetting is the principle of linearity with divisibility. Since your gcd divides both a+b and a-b, it is then also true that your gcd will divide ANY linear combination of these integers (assuming integer weights, of course).

Namely, instead of being completely general with divisibility, realize that you can manipulate the variables x and y in the expression d | (a+b)x + (a-b)y to get a form that is equally valid. I think that your book wants you to assume that a and b are coprime integers, as well.

So, why not try playing around and actually picking numerical values for x and y that can help eliminate either a or b?