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Taking the Fifth (Posted on 2014-06-21) Difficulty: 3 of 5

For the most part, if z^5 is the sum of two positive cubes, then z^2 is itself the sum of positive cubes:
(z^2 = x^3 + y^3, so z^2*z^3 = z^5 = (z*x)^3 + (z*y)^3).

Hence, such solutions to z^5=x^3+y^3 are said to be 'trivial'

In this sense, 3549^5 is the non-trivial sum of two positive cubes. It is nevertheless possible to calculate the cubes without resort to brute force.

How, and what are they?

No Solution Yet Submitted by broll    
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Hints/Tips Another clue | Comment 2 of 5 |

3549^5 = 476749^3+768950^3

Now, how was it done?


  Posted by broll on 2014-07-04 08:46:47
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