13

The sequence A023042 on the OEIS website shows that a large percentage of $N^3$ are a sum of three positive cubes. OEIS lists only N<1770, but we can extend that:$$\begin{array}{|c|c|} N&\text{%}\\ 2000&85.8\text{%}\\ 4000&89.8\text{%}\\ 6000&92.1\text{%}\\ 8000&93.3\text{%}\\ 10000&94.2\text{%}\\ \end{array}$$

This means that 94.2% of all N<10000 have a solution to $a^3+b^3+c^3=N^3$ in positive integers. Thus, if we pick a random N in the high end of that range, there is a very good chance that there is an a,b,c.

Now my question is that:Are there infinitely many $N^3$ (especially for prime $N$) that cannot be expressed as a sum of three positive cubes?

For example, there are no positive integers,

$a^3+b^3+c^3=999959^3$

even though the percentage of $N<1000000$ with solutions should be close to 99%

The link to the OEIS website is:https://oeis.org/A023042

This question was originally posted here:https://math.stackexchange.com/questions/1120136/are-there-infinitely-many-n3-especially-for-prime-n-that-cannot-be-expres but hasn't been answered yet,so I have posted it here

Community
  • 1
Remember me
  • 181
  • 1
    This problem seems quite difficult. The issue, of course, is that there are no local obstructions in this case: there are always solutions in all $\mathbb{Z}_p$ and $\mathbb{R}$. So you would need to try to use a finer obstruction, e.g. a Brauer-Manin obstruction. – Daniel Loughran May 31 '15 at 08:29
  • 2
    Also, as you no doubt know, determining which integers are a sum of three (possibly negative) cubes is a very difficult open problem (see e.g. http://mathoverflow.net/questions/138886/which-integers-can-be-expressed-as-a-sum-of-three-cubes-in-infinitely-many-ways). One would think that your problem were easier as you are working with a cube, but restricting to positive cubes seems to complicate things. – Daniel Loughran May 31 '15 at 08:46

0 Answers0