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From: leimkuhl@uiucdcsp.CS.UIUC.EDU
Newsgroups: net.math
Subject: Re: A real problem! (Not a polar be
Message-ID: <9600030@uiucdcsp>
Date: Sun, 17-Nov-85 17:43:00 EST
Article-I.D.: uiucdcsp.9600030
Posted: Sun Nov 17 17:43:00 1985
Date-Received: Tue, 19-Nov-85 04:02:31 EST
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Nf-From: uiucdcsp.CS.UIUC.EDU!leimkuhl    Nov 17 16:43:00 1985




bs is correct in his analysis, but he did not need to break the
problem down to cases.

We seek solutions in integers to n^2 + 2n - 3 = (r^2)/3.
Obviously this means r is divisible by three--r=3q.  So r^2=9q^2,
and we can replace our first equation by n^2 +2n - 3 = 3q^2.

Now we can just complete the square:

	q^2 + 2q +1 = (q+1)^2 = 3r^2 + 4

Which is the same equation bs derives via his case by case analysis.

-Ben Leimkuhler