find the gcd(a+b,a-b)
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I don't know bro
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I want to show that for gcd(a,b)=1 a,b∈Z gcd(a+b,a−b)=1 or gcd(a+b,a−b)=2 holds.
I think the first step should look something like this:
d=gcd(a+b,a−b)=gcd(2a,a−b)
From here I tried to proceed with two cases.
1: a−b is even, which leads to gcd(a+b,a−b)=2
2: a−b is odd, which leads to gcd(a+b,a−b)=1
My main problem I think is, that I do not know how I should include gcd(a,b)=1 in the proof.
Any help is appreciated. Thx in advance.
Cherio Woltan
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