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Geometric Integers II (Posted on 2011-08-02) Difficulty: 3 of 5
Consider three positive integers x, y and z in geometric sequence with x < y < z < 2011 and, x+y=z+1 and determine all possible triplets (x,y,z) that satisfy the given conditions.

No Solution Yet Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

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Solution Fibonacci | Comment 2 of 3 |
The values of x, y, and z can be found from Fibonacci numbers.
x=fibonacci(2n-1)^2
y=fibonacci(2n-1)*fibonacci(2n)
z=fibonacci(2n)^2
x<y<z<2011 for n=1, 2, 3, 4.
n   x   y   z
1   1   1   1
2   4   6   9
3   25  40  64
4   169 273 441
Therefore, the possible triplets are (1, 1, 1), (4, 6, 9), (25, 40, 64), and (169, 273, 441).



  Posted by Math Man on 2011-08-03 20:34:14
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