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 Taking the Fifth (Posted on 2014-06-21)

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 No Rating

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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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