TITLE: What's special about the greatest common divisor of a + b and a - b?
QUESTION [5 upvotes]: Problem:

Prove that if gcd( a, b ) = 1, then gcd( a - b, a + b ) is either 1 or 2.

From Bezout's Theorem, I see that am + bn = 1, and a, b are relative primes. However, I could not find a way to link this idea to a - b and a + b. I realized that in order to have gcd( a, b ) = 1, they must not be both even. I played around with some examples (13, 17), ...and I saw it's actually true :( ! Any idea?

REPLY [2 votes]: Hint $\rm\,\ a\!-\!b + (a\!+\!b)\ {\it i}\ =\ (1\!+\!{\it i})\ (a\!+\!b\!\ {\it i})\ \ $ yields a slick proof using Gaussian integers. This reveals the arithmetical essence of the matter and, hence, suggests obvious generalizations.