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Geometric Integers II (Posted on 2011-08-02)

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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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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