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 Some Squares Sum Square (Posted on 2012-07-30)
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

 No Solution Yet Submitted by K Sengupta No Rating

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 re: Solution (spoiler) | Comment 7 of 9 |
(In reply to Solution by Harry)

Having now read the hint from KS, I’m wondering if this is what was needed:

The given equation can be rearranged to

10(2G – 3F)2 – 18(2G – 3F)(H – 3F) + 13(H – 3F)2 = 0                    (1)

Clearly, a solution is 3F = 2G = H, which gives triples 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 triple is (2, 3, 6).

___________

No other triple is possible. Writing (1) as 10a2 – 18ab + 13b2 = 0 and solving

this quadratic for a/b would involve the discriminant (-18)2 – 4(10)(13),

which is negative and would deny real values for at least one of a and b and

therefore at least one of F, G and H.

 Posted by Harry on 2012-08-01 14:06:37

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