I already know that gcd(a,b)=gcd(a-b,b). However, can I also say that gcd(a,b)=gcd(a+b,b)? I think this is correct and the proof is simple: consider gcd(a+b,b). Then apply this subtraction property of GCD's to say gcd(a+b,b)=gcd(a+b-b,b)=gcd(a,b) as required.
2026-04-02 13:56:54.1775138214
Addition property for GCD
32 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtRelated Questions in GCD-AND-LCM
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