Re: Diophantine equation

From: Ignacio Larrosa Cañestro (ilarrosaQUITARMAYUSCULAS_at_mundo-r.com)
Date: 01/30/05 

Date: Sun, 30 Jan 2005 14:44:20 +0100

En el mensaje:2022-41FCD0B1-6@storefull-3252.bay.webtv.net,
Oscar Lanzi III <ol3@webtv.net> escribió:
> 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)

Hera you has lost a 2. It must be

k - 1 = 2(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 it must be

m^2 - 3n^2 <= 8

But, taking in account the former typo, it is

2(2n^2+4) >= (m^2-n^2-4)

12 >= m^2 - 5n^2

> 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.

Actually, there are more solutions (2, 1), (3, 1), (9, 2), (14, 3), (43, 9),
(67, 14), ... . For all of them, k = 5. 

-- 
Best regards,
Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosaQUITARMAYUSCULAS@mundo-r.com