I started by saying that gcd(a,b)=d1 and gcd(a+b,gcd(a,b))=d2
Then I tried to show that d1≥d2,d1≤d2.
I know that d2|gcd(a+b,d1) hence d2≤d1.
How do I prove that d2≥d1 ?
Answer
If gcd(a,b)=d1 then a=d1x and b=d1y, where x,y are integers. Consequently,
gcd(a+b,gcd(a,b))=gcd(d1(x+y),d1)=d1gcd(x+y,1)=d1.
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