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A multi-solution diophantine problem (Posted on 2006-09-15) Difficulty: 3 of 5
Consider the equation x^2+y^5=z^3 where x, y, and z, are positive integers.

(A) Can you give at least three solutions to it?
(B) Determine whether or not there is an infinite number of solutions.

  Submitted by K Sengupta    
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Solution: (Hide)
Substituting: x = 2a, y = 2b, and z = 2c, in the original equation, one obtains:
2^2a + 2^5b = 2^3c

Considering 2^2a + 2^5b = 2^3c in conjunction with the identity
2^(k-1) + 2^(k-1) = 2^k
generates the equation :
2a + 1 = 5b + 1 = 3c

Since, the minimum possible positive integral solution to (#) is (a,b,c)=(10,4,7), by way of application of Chinese Remainder Theorem to (#), we obtain the following parametric relationship:
(a,b,c) = (15*s + 10, 6*s + 4, 10*s + 7)---------(#), so that
(#) generates infinite number of triplets (a,b,c) corresponding to integral s (positive 0r negative)

Since, by assumption x=2^a; y=2^b and z = 2^c; it immediately follows that there exists an infinite number of solutions to the equation under reference. For example, for s= 1, 2, 3, 4, 5, 6 and 7 we obtain seven triplets:

(x,y,z) = (2^10, 2^4, 2^7), ( 2^25, 2^10, 2^17), (2^40, 2^16, 2^27), (2^55, 2^22, 2^37), ( 2^70, 2^28, 2^47), (2^85, 2^34, 2^57), (2^100, 2^40, 2^67)

NOTE: The parametric relationship in (#) may correspond to one of the numerous parametric solutions for the problem under reference. However, the relationship (#) alone is sufficient to prove that the total number of solutions to the problem is indeed infinite.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re: No SubjectK Sengupta2006-12-31 12:43:13
No SubjectFerdinand2006-12-31 07:27:48
Some Thoughtsre: re: no subjectDennis2006-09-17 17:59:06
re: No SubjectJohn Reid2006-09-15 15:19:21
Some ThoughtsNo SubjectDennis2006-09-15 14:30:07
SolutionSolution for part ADej Mar2006-09-15 13:15:57
SolutionMany solutions (all of them?)Old Original Oskar!2006-09-15 11:49:40
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