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 March 5th, 2009, 06:18 AM #1 Member   Joined: Nov 2006 Posts: 54 Thanks: 0 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). 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? March 5th, 2009, 08:10 AM   #3
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Re: A Harmonic And Near Pythagorean Puzzle

Quote:
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. March 5th, 2009, 08:49 AM #4 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: A Harmonic And Near Pythagorean Puzzle (3, 4, 6) was the only solution I could find. There aren't any other small solutions. Tags harmonic, puzzle, pythagorean pythagorean puzzle

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