I was intrigued that the previous computer posts could find only two solutions for positive values of a, b and c, so took to pen and paper:

(a + b)^{2}= a^{2} + b^{2} + 2ab= c^{2} + 2ab(1)

(1000 – a)(1000 – b) = (b + c)(c + a) = c(a + b + c) + ab = 1000c + ab

= 1000c + [(a + b)^{2} – c^{2}]/2using (1)

= 1000c + [(1000 – c)^{2} – c^{2}]/2

(1000 – a)(1000 – b)= 500000(2)

500000 = (2^{5})(5^{6}), so its factors must all be of the form 2^{x}5^{y} where x and y

range from 0 to 5 and 6 respectively, giving 6*7 = 42 ordered pairs as follows. (1,500000),....,(500,1000), (625,800), (800,625),....,(1000,500),(500000,1).

There are also 42 corresponding pairs of negative factors, giving a total of 84 solutions to equation (2). If the two brackets in (2) are positive then a and b must both be less than 1000, so the only factors possible are the underlined central ones, giving 1000 – a = 625 and 1000 – b = 800 from which (a,b,c) = (375,200,425) or 1000 – a = 800 and 1000 – b = 625 from which (a,b,c) = (200,375,425).

The other 82 possible pairs give solutions involving negative values for one of a, b and c, as listed in previous posts.