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March 5th, 2009, 06:18 AM   #1
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A Harmonic And Near Pythagorean Puzzle

Determine all possible triplet(s) (x,y,z) of positive integers, with x< y< z, such that:
x,y and z (in this order) are in harmonic sequence, and: x^2 + y^2 = z^2 - 11

Note: x, y and z (in this order) are in harmonic progression (or, harmonic sequence), if 1/x, 1/y and 1/z(in this order) are in arithmetic progression (or, arithmetic sequence).
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March 5th, 2009, 08:01 AM   #2
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Re: A Harmonic And Near Pythagorean Puzzle

Quote:
Originally Posted by K Sengupta
x,y and z (in this order) are in harmonic sequence
What does this condition mean? That y is the harmonic mean of x and z?
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March 5th, 2009, 08:10 AM   #3
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Re: A Harmonic And Near Pythagorean Puzzle

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Originally Posted by CRGreathouse
Quote:
Originally Posted by K Sengupta
x,y and z (in this order) are in harmonic sequence
What does this condition mean? That y is the harmonic mean of x and z?
Yes.

x, y and z are in harmonic progression (or, harmonic sequence), if 1/x, 1/y and 1/z are in arithmetic progression (or, arithmetic sequence). So, y is the harmonic mean of x and z.

I confirm having amended the original problem text to make the intent clearer.
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March 5th, 2009, 08:49 AM   #4
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Re: A Harmonic And Near Pythagorean Puzzle

(3, 4, 6) was the only solution I could find. There aren't any other small solutions.
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