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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.

Bill Dubuque
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user124820929
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    Yes. $\gcd(a,b) = \gcd(a \pm kb, b)$ for any non-zero integer $k$ and $\gcd(a,b) = \gcd(a, b \pm ja)$ for any non-zero integer $j$. But don't make the mistake of skipping a step and assuming $\gcd(a,b) = \gcd(a +kb, ja + b)$. That isn't true. – fleablood Jan 25 '24 at 01:34

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