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Re: the missing bit » linkadge

Posted by AuntieMel on January 16, 2007, at 9:17:37

In reply to Anyone good at number theory?, posted by linkadge on January 11, 2007, at 17:17:02


I found a property:

GCD (a + b, b) = GCD (a, b)

So GCD (a^2 + b^2, a + b) = GCD (a^2, a + b).

The only possible cases for a and b are
1) a is odd, b is even
2) a is even, b is odd
3) a and b are both odd

In case 1, the GCD of (a^2, a + b) is 1
case 2, the GCD of (a^2, a + b) is 1 (a + b is odd)
case 3, the GCD of (a^2, a + b) is 2 (a + b is even)

 

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poster:AuntieMel thread:721428
URL: http://www.dr-bob.org/babble/social/20070112/msgs/722827.html