(In reply to
Start by armando)
Given:A^3+B^3 = 31C^3
If (36*31y)^3+ (36*31+y)^3 = 31*(6x)^3, then x^3y^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(pq)/(p+q). y =(1116(pq))/(p+q), n=31
Checking:(372/(p+q))^3 (1245456 (pq)^2)/(p+q)^2= 415152
q = (1)^(2/3) (31p^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) (31p^3)^(1/3), and p=137/42, as expected.
Edited on April 23, 2016, 10:27 pm

Posted by broll
on 20160422 23:36:04 