I found an amazing conjecture: the cube of every perfect number can be written as the sum of three positive cubes. The equation is
$$x^3+y^3+z^3=\sigma^3$$
where $\sigma$ is a perfect number
(well it holds good for the first three perfect numbers that is):
$${ 3 }^{ 3 }+{ 4 }^{ 3 }+{ 5 }^{ 3 }={ 6 }^{ 3 }$$
$${ 21 }^{ 3 }+{ 18 }^{ 3 }+{ 19 }^{ 3 }={ 28 }^{ 3 }$$
$${ 495 }^{ 3 }+{ 82 }^{ 3 }+{ 57 }^{ 3 }={ 496 }^{ 3 }$$
Is this what I am proposing that the cube of any perfect number can be expressed in terms of the sum of three positive cubes true?
If it is then can we prove it?
Best Answer
As pointed out by E. Schmidt, the sequence A023042 shows that a large percentage of cubes $N^3$ are a sum of three positive cubes. OEIS lists only $N<1770$, but we can extend that:
$$\begin{array}{|c|c|} \hline N&\text{%}\\ \hline 2000&85.8\text{%}\\ 4000&89.8\text{%}\\ 6000&92.1\text{%}\\ 8000&93.3\text{%}\\ 10000&94.2\text{%}\\ \hline \end{array}$$
This means that $94.2\text{%}$ of all $N<10000$ have a solution to $a^3+b^3+c^3=N^3$ in positive integers. Note that $N=10000$ is still small. Extrapolating the table, it can be seen that the percentage may easily reach $99\text{%}$ if we go into the millions.
Thus, if we pick a random $N$ in the high end of the range, there is a very good chance that there is an $a,b,c$. For the next perfect number $N=8128$, it is just mere statistics that suggests $N^3$ will be the sum of three positive cubes, and not because it is perfect. In fact, like $496$, it is in several ways,
$$2979^3 + 4005^3 + 7642^3 = 8128^3$$
$$2^6(102^3 + 673^3 + 2007^3) = 8128^3$$
$$2^9(197^3 + 198^3 + 1011^3) = 8128^3$$
And it was almost certain for the next perfect number which is in the millions,
$$2^{27}(3042^3 + 56979^3 + 45845^3) = 33550336^3$$
$$2^{30}(821^3 + 32590^3 + 8227^3) = 33550336^3$$
$$2^{36}(4543^3 + 6860^3 + 5104^3) = 33550336^3$$
Both can be expressed in many more ways than this, and I have only chosen a sample. For the cube of the next perfect number, or $137438691328^3$, chances are even greater that it is a sum of three positive cubes in many ways as well.
Update: Yes, it is:
$$2^{54}(425664^3 + 358719^3 + 275140^3)= 137438691328^3$$
$$2^{54}(432204^3 + 386604^3 + 177535^3)= 137438691328^3$$
Note: Jarek Wroblewski has calculated $a^3+b^3+c^3 = d^3$ with $\color{brown}{\text{co-prime}}$ $a,b,c$, and $d<1000000$ in his website. Using his database and some help with Mathematica and Excel, I came up with the table above which counts all $N$, regardless of whether $a,b,c$ is co-prime or not.
P.S: An interesting question, I believe, is: "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 $99\text{%}$.