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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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Solution Solution | Comment 1 of 3
If 81 + K and 162 + K are both perfect squares, say, n^2 and m^2, then we can write m^2 - n^2 = (162 + K - 81 - K) = 81 = 9^2
m^2 - n^2 = (m+n)(m-n) and 81 can be factored into two pieces in only three ways: 1x81, 3x27, and 9x9. 9x9 we can discard immediately since we know n > 0. [If n = 0 then K = -162 and 81+K is negative and hence not a perfect square.]

If (m+n)(m-n) = 3*27 then m = 15 and n = 12 and K = 63, since 63 + 81 = 12^2 and 63 + 162 = 15^2.

If (m+n)(m-n) = 1*81 then m = 40 and n = 1 and K = 1519 similarly.

If K = 1519, though, then 337+K = 1856 which is not a perfect square so that's not a valid option.

If K = 63, then 337+K = 400 = 20^2 so this value of K does yield 3 perfect squares. Are they a harmonic sequence? Yes, since 1/12 = 5/60, 1/15 = 4/60 , and 1/20 = 3/60 and these inverses are an arithmetic sequence. So K = 63 is a solution, and no other value of K meets these requirements.
  Posted by Paul on 2013-11-13 19:58:55
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