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Some Squares Sum Square (Posted on 2012-07-30) Difficulty: 2 of 5
Determine all possible non zero real triplet(s) (F, G, H) that satisfy the following system of equations:

(7F)2 + (7G) 2 + (7H) 2 = (2F + 3G + 6H) 2 , and :

1/F + 1/G + 1/H = 1

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(3): Got it (spoiler) Comment 9 of 9 |
(In reply to re(2): Solution (spoiler) by K Sengupta)

Aha!

So, after much helpful advice...

First rearrange the given equation to

                        4(3F – H)2 + (3F - 2G)2 + 9(H – 2G)2 = 0

then deduce that
3F = 2G = H, so any triple must be of the form (f, 3f/2, 3f).

Substitution into the equation 1/F + 1/G + 1/H = 1 then gives

f = 1 + 2/3 + 1/3  =  2, so the only possible triple is (2, 3, 6).



  Posted by Harry on 2012-08-02 10:51:23

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