Re: Diophantine equation

From: Oscar Lanzi III (ol3_at_webtv.net)
Date: 01/30/05 

Date: Sun, 30 Jan 2005 06:18:57 -0600

Assume wlog that a >= b. Set m = a+b and n = a-b, where perforce m and
n are both even or both odd because of the difference being 2b. Also m
and n are both nonnegative with a >= b. Then a^2+b^2 = (m^2+n^2)/2
and ab = (m^2-n^2)/4. Thus

k = (a^2+b^2) /(ab-1)
   = 2 (m^2+n^2)/(m^2-n^2-4).

where k is a positive integer. Subtract 1 from both sides:

k - 1 = (2n^2+4)/(m^2-n^2-4)

Now k can't be 1 because the numerator 2n^2+4 is identically nonzero, so
k > = 2 and k-1 >= 1. Ergo 2n^2+4 >= m^2-n^2-4 and m^2-n^2 <= 8. But
m-n = 2b >= 2 so m+n <= 8/2 = 4. We are forced to accept m = 3, n = 1
as the only posible solution, corresponding to a = 2, b = 1.

--OL

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