Harmonic Sum Of Squares (Posted on 2006-01-31)

	Submitted by K Sengupta
Rating: 2.7500 (4 votes)
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At the outset, I would wish to acknowledge the accurate methodology employed by goFish culminating in an accurate solution to the problem under reference. SOLUTION TO THE PROBLEM: X = 493; Y = 697 and Z = 1189 constitute the only solution to the problem under reference. EXPLANATION: Since X,Y and Z are in H.P. with X < Y < Z , we must have Y = 2*X*Z/(X+Z). Accordingly, X + Y = Z + 1 gives 2XZ/(X+Z) = Z - X + 1 or, p^2-(2*Z)^2=P, where P = X+Z or, (2P-1)^2 = 2*(2Z)^2 + 1 ........(#) In terms of Pell's Equation, we observe that 2*(2Z)^2 + 1 is a perfect square for Z < 2006 whenever Z = 1,6,35,204 and 1189. Now, only Z=1189 is expessible as the sum of two distinct squares since 1189 = 10^2 + 33^2. None of the four other values of Z are expressible as sum of squares of two distinct positive integers. Consequently, Z=1189, giving (2P-1) = 3363 or, P = 1682, so that X = 1682 - 1189 = 493 Hence, Y = 2*493*1189/(493 + 1189) = 697. Now, 493 = 22^2 + 3^2 and 697 = 24^2 + 11^2 , which is in conformity with provisions of the problem. Consequently, X= 493, Y= 697 and Z = 1189 constitutes the only solution to the problem under reference.

	Subject	Author	Date
	Solution - one way	goFish	2006-01-31 17:15:12
	re: Consecutive	Gamer	2006-01-31 17:11:35
	Consecutive	owl	2006-01-31 16:37:11