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 } ] .

#NumberTheory

Easy Math Editor

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Comments

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.

Calvin Lin Staff - 4 years, 11 months ago

Yes, it's solved :)

D K - 4 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$