Prove that \(\gcd(a+b, a-b) \geq \gcd(a,b) \), where \(a\) and \(b\) are two integers.

Probable direction: Lowest positive value of the equation {ax + by} will give the h.c.f of a & b.

While the lowest positive value { (a+b)x+(a-b)y } will give the h.c.f. of (a+b) & (a-b).

[ x and y are integer variables]

So, the proof comes down to : [the lowest positive value of { (a+b)x + (a-b)y } ] >= [ lowest positive value of { ax + by } ] .

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TopNewestGreat! Do you now see why the last step is obvious?

Hint:Do not let your notation do double duty. It would be helpful to use \(x_1, y_1 \) in one of them, and \( x_2, y_2 \) in the other. This way, you can ask about the relationship of these terms. – Calvin Lin Staff · 11 months, 1 week agoLog in to reply

– D K · 11 months ago

Yes, it's solved :)Log in to reply