# Problem Regarding H.C.F.

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 } ] .

Note by D K
2 years ago

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Great! 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.

Staff - 1 year, 12 months ago

Yes, it's solved :)

- 1 year, 12 months ago