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Perfectly Harmonic (Posted on 2013-11-13) Difficulty: 3 of 5
Each of the three positive integers (81+K), (162+K) and (337+K) is a perfect square such that their positive square-roots are in harmonic sequence.

Determine the value of K, and prove that no other value for K is possible.

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

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re: Solution | Comment 2 of 3 |
(In reply to Solution by Paul)

If n=0, then K=-81, so K+81=0 is a square. However, I got the same solution:K=63. Then, K+81=144=12^2, K+162=225=15^2, and K+337=400=20^2.

  Posted by Math Man on 2013-11-14 20:26:40
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