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 A Pythagorean triplet (Posted on 2012-08-28)
Given that a2+b2 = c2
find the value of a*b*c if a+b+c=1000.

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 Pen & paper approach (spoiler) Comment 4 of 4 |
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 = a2 + b2 + 2ab  = c2 + 2ab                                             (1)

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

= 1000c + [(a + b)2 – c2]/2         using (1)

= 1000c + [(1000 – c)2 – c2]/2

(1000 – a)(1000 – b)      = 500000                                              (2)

500000 = (25)(56), so its factors must all be of the form 2x5y 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.

 Posted by Harry on 2012-08-30 19:18:15

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