 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  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 Comments: ( Back to comment list | You must be logged in to post comments.) re(2): Solution Comment 5 of 5 | (In reply to re: Solution by broll)

I compared the two sequences A051394 and A050801.  The OEIS has a file with the first 185 terms of A051394 and 1000 terms of A050801.

62500 is the 185th term of A051394 but only the 167th term of A050801, meaning that there are 18 nontrivial values less than 62500.  Comparing those lists I found 3549, 4914, 5054, 5526, 10545, 20850, 24087, 24510, 28392, 32550, 38532, 38829, 45612, 48734, 56316, 58632, 62034, 62465 are the first 18 values in A051394 but not in A050801.

 Posted by Brian Smith on 2017-05-05 23:54:15 Please log in:

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