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Cubic Diophantine Conclusion (Posted on 2016-04-22) Difficulty: 3 of 5
Each of A, B and C is a positive integer that satisfies this equation:

A3 + B3 = 31C3

Find the smallest value of A+B+C

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts re: Start | Comment 3 of 9 |
(In reply to Start by armando)

Given:A^3+B^3 = 31C^3

If (36*31-y)^3+ (36*31+y)^3 =  31*(6x)^3, then x^3-y^2 = 415152
This has a known solution < 100000: x=217, y=3131 (or -3131)
Let x=12n/(p+q),  x=372/(p+q),  n=31
y = -36n(p-q)/(p+q). y =-(1116(p-q))/(p+q), n=31
Checking:(372/(p+q))^3- (1245456 (p-q)^2)/(p+q)^2= 415152
q = (-1)^(2/3) (31-p^3)^(1/3)
Plugging in the known solution:
giving: p=137/42, q=-65/42; A=137, B=-65, C=42 is a solution.
Checking again: -65/42 = (-1)^(2/3) (31-p^3)^(1/3), and p=137/42, as expected.


Edited on April 23, 2016, 10:27 pm
  Posted by broll on 2016-04-22 23:36:04

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