A Cubic And Biquadratic Problem (Posted on 2006-06-18)

	Submitted by K Sengupta
Rating: 3.6667 (3 votes)
Solution:	(Hide)
The three smallest numbers are (625,1250,125) in this order EXPLANATION: Let the three numbers be A, B and D in this order. Then from provisions of the problem, we obtain: A^2 + B^2 = D^3 and A^3 + B^3 = 9*(D^4) It is given that B > A So, A^2 + B^2 = D^3 or, p^2 + q^2 = D, where A = p*D and B = q*D, for two positive integers p and q. This is only feasible if there exist positive integers b, T1 and T2 such that p = C*T1 and q = C*T2, where T1 and T2 must be coprime.-------------(!#) Accordingly; A = p(p^2 + q^2); B = q(p^2 + q^2) and D = p^2 + q^2. Substituting these values in the relationship A^3 + B^3 = 9*(D^4); we obtain: p^3 + q^3 = 9(p^2 + q^2) or, C(T1^3 + T2^3) = 9(T1^2 + T2^2); where p=C*T1 and q = C*T2 Since B is greater than A , we must have q > p, so that T2 > T1 . Since the minimum value for which T1 and T2 are coprime in conformity with condition (!#)occurs at (T1, T2)= (1,2) ; substituting T1=1 and T2=2, we obtain: C= 9*(2^2 + 1^2)/(2^3+ 1^3) = 5, giving p=5 and q = 10, so that: A = p(p^2+q^2) = 625 B = q(p^2+q^2) = 1250 and D = p(p^2+q^2) = 125 Hence, the three smallest numbers are (625,1250,125) in this order.

	Subject	Author	Date
	computer solution (spoiler)	Charlie	2006-06-18 13:22:25