Given that a and b are relatively prime integers. Then gcd(a, b)=1 or in equivalent notation (a, b)=1.
Let d=(a+2b, a−2b). Then it follows that d∣(a+2b) and d∣(a−2b).
That is d∣(m(a+2b)+n(a−2b)),
2(a+2b)+2(a−2b)=4a and thus d∣4a,
2(a+2b)−2(a−2b)=4b and thus d∣4b.
Hence d∣(4a, 4b). We have that d∣4.
We can rule out every possibility except d=1,2 or d=4. However, we do not yet know which of those three values are actually possible.
Let's show that all possible values of d∈{1,2,4} actually occur:
a=1,b=0 gives d=1,
a=0,b=1 gives d=2,
a=2,b=1 gives d=4.
Therefore, the answer is d∈{1,2,4}.
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