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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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re: Solution | Comment 4 of 5 |
(In reply to Solution by Brian Smith)

Gosh, it's so long since I set this that my workings are in my old computer (now sadly deceased)!

Thanks for tackling it!

Update: I've now found part of it in a folder I kept. The insight I used was that if a solution is not of the 'trivial' form then it is (always? probably? - not clear from my notes) of the form z^5=a^3(b^3+c^3).

When a = 7*13^3, then (2*5^2)^3(7*13^3)^3+31^3(7*13^3)^3.

Your line  7^2*13^6*(4*7^2*13-3^7) hints at a somewhat similar relationship. I'll look further and see if I can find the rest of the working that covers the missing points. The inspiration for the problem came from A051394, because if 3549^5 is a 'non-trivial' sum of cubes, then what are the cubes?


Edited on May 5, 2017, 10:57 am
  Posted by broll on 2017-05-05 09:20:26

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