Let \(d=(a+b, a-b)\), then:
$$ d|a+b $$$$ d|a-b $$
According to this lemma, \(d\) can divide any linear combination of \((a+b)\) and \((a-b)\):
$$ d | (a+b) + (a-b) \implies d | 2a $$$$ d | (a+b) - (a-b) \implies d | 2b $$
Since \(d\) divides both \(2a\) and \(2b\), and since every linear combination of \(2a\) and \(2b\) is a multiple of \((2a, 2b)\), then:
$$ d|(2a, 2b) $$
According to this lemma, we can take the 2 out:
$$ (a, b) = 1 \implies (2a, 2b)=2 $$$$ \therefore d|2$$
This means \(d\) can only be 2 or 1.