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.

(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