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Topic: What squares are the sums of two or more cubes?
Replies: 6   Last Post: Oct 19, 2012 9:18 PM

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James Waldby

Posts: 347
Registered: 1/27/11
Re: What squares are the sums of two or more cubes?
Posted: Oct 18, 2012 9:57 PM
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On Fri, 19 Oct 2012 01:06:32 +0100, Peter Percival wrote:

> What squares of positive integers are the sums of two or more distinct
> cubes of positive integers?
>
> 9 = 1 + 8, 36 = 1 + 8 + 27, 100 = 1 + 8 + 27 + 64, 225 = 1 + 8 + 27 + 64 +
> 125, etc.
>
> On the LHSs are the squares of the triangular numbers. Are there any
> examples not in that sequence?


I don't know which sequence you mean, but note that Nicomachus's
theorem (first century CE) covers the cases you mentioned. See
<http://en.wikipedia.org/wiki/Squared_triangular_number>

Liouville (1809 ? 1882) generalized to cubed sums of divisor counts.
<http://amca01.wordpress.com/2011/01/10/a-cute-result-relating-to-sums-of-cubes/>

As noted in <http://mathworld.wolfram.com/CubicNumber.html>, every
positive number is a sum of no more than 9 positive cubes, "proved by
Dickson, Pillai, and Niven in the early twentieth century", and every
sufficiently large integer is a sum of no more than 7 positive cubes.
"Deshouillers et al. (2000) conjectured that 7373170279850 is the
largest integer that cannot be expressed as the sum of four nonnegative
cubes".

Let S be the set of sums of three cubes of numbers that are in the
range 0 to n. There are n^3 such sums. Even though many of the sums
have duplicated values, S nevertheless covers on the order of n^3
values in the range from 0 to 3 n^3, so naively the chance of a
number being the sum of three cubes is in the neighborhood of 1/3.
Let T be the set of sums of four cubes of numbers that are in the
range 0 to n. This gives some fraction of n^4 distinct sums for
numbers in the range 0 to 4 n^3, so on the average there are in the
neighborhood of n/4 ways to write a number less than 4 n^3 as the
sum of 4 cubes. However, among even small squares it looks like
14^2, 18^2, 20^2, and 26^2 require five cubes in their sums:

0 : 0 = 0 + 0 + 0 + 0 + 0
2 : 4 = 0 + 1 + 1 + 1 + 1
4 : 16 = 0 + 0 + 0 + 8 + 8
6 : 36 = 0 + 0 + 1 + 8 + 27
8 : 64 = 0 + 0 + 0 + 0 + 64
8 : 64 = 1 + 1 + 8 + 27 + 27
10 : 100 = 0 + 1 + 8 + 27 + 64
12 : 144 = 0 + 8 + 8 + 64 + 64
16 : 256 = 0 + 64 + 64 + 64 + 64
18 : 324 = 27 + 27 + 27 + 27 + 216
22 : 484 = 0 + 8 + 8 + 125 + 343
24 : 576 = 0 + 0 + 0 + 64 + 512
24 : 576 = 1 + 8 + 8 + 216 + 343
26 : 676 = 1 + 27 + 216 + 216 + 216
28 : 784 = 0 + 1 + 27 + 27 + 729
28 : 784 = 1 + 8 + 216 + 216 + 343
30 : 900 = 0 + 125 + 216 + 216 + 343
32 : 1024 = 0 + 0 + 0 + 512 + 512

--
jiw



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