Posted by Dr. Bob on January 17, 2007, at 1:06:50
In reply to Re: Well, the answer..., posted by linkadge on January 12, 2007, at 16:54:19
> GCD (a + b, b) = GCD (a, b)
>
> So GCD (a^2 + b^2, a + b) = GCD (a^2, a + b).I don't know, the first would seem to imply:
GCD (a^2 + a + b, a + b) = GCD (a^2, a + b)
> a^2 + b^2 doesn't factor into (a+b)(a-b).
I think that might be the direction to go, maybe think of it as:
GCD ((a + b)^2 - 2ab, a + b)
Bob
poster:Dr. Bob
thread:721428
URL: http://www.dr-bob.org/babble/social/20070112/msgs/723064.html