For example (1): supose gcd(2a + b, 3a + 2b) = d ==> there exist integers n, m s.t.

3a + 2b = md

2a + b = nd --- rest these two equations an get

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a + b = (m-n)d --- now rest this from the the second eq. above:

a = (m + 2n)d

But then d divides a and d divides 2a + b ==> d divides b ==> d = 1 .

Tonio