Can the cube of every perfect number be written as the sum of three cubes?

Question

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?

Answer 1

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:

$$2^{54}(425664^3 + 358719^3 + 275140^3)= 137438691328^3$$

$$2^{54}(432204^3 + 386604^3 + 177535^3)= 137438691328^3$$

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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 $99\text{%}$.

Answer 2

Well, I played myself with this formula and proved that if the following ratio holds:

$$\frac{a^3 + b^3 + c^3}{abc} = 6$$

then the three integers:

$$x_1 = a + b$$ $$x_2 = a + c$$ $$x_3 = b + c$$

are such that: $$ x_1^3 + x_2^3 + x_3^3 = y^3 $$

Proof is very simple:

Compute: $$ (a+b+c)^3$$

this gives: $$a^3 + b^3 + c^3 + 3[a^2(b+c) + b^2(a+c) + c^2(a+b)] + 6abc$$ now gather these addendums in three variables, forgetting about "6abc" for a while:

$$\alpha = a^3 + 3a^2b + 3ab^2$$ $$\beta = b^3 + 3b^2c + 3bc^2$$ $$\gamma = c^3 + 3c^2a + 3ca^2$$

now complete the gathering, adding to each variable a fraction of "6abc", obtaining:

$$x_1^3 = a^3 + 3a^2b + 3ab^2 + k_1abc$$ $$x_2^3 = b^3 + 3b^2c + 3bc^2 + k_2abc$$ $$x_3^3 = c^3 + 3c^2a + 3ca^2 + k_3abc$$

these cubes are perfect if and only if: $$k_1abc = b^3$$ $$k_2abc = c^3$$ $$k_3abc = a^3$$

with the additional condition: $$k_1 + k_2 + k_3 = 6$$ where $$ k_1, k_2, k_3$$ are real Now, the following equivalence holds:

$$a^3+b^3+c^3=6abc$$

and so:

$$\frac{a^3+b^3+c^3}{abc}=6$$

when this holds, the three numbers become: $$x_1^3 = (a+b)^3$$ $$x_2^3 = (b+c)^3$$ $$x_3^3 = (c+a)^3$$

and then we have our sum of cubes.

Aftermaths!

This procedure does not say how to find the three numbers, but indeed once you find them, it is easy to show that each group of three base numbers, multiplied by an integer "h" still gives a sum of cubes that results in a perfect cube. As an example, say: (a, b, c) = (1, 2, 3) then: $$x_1 = 1 + 2 = 3$$ $$x_2 = 2 + 3 = 5$$ $$x_3 = 3 + 1 = 4$$

but also:

$$(a, b, c) = h(1, 2, 3)$$ with $$h \in \mathbb R$$ is a solution. Proof is starightforward:

if $$\frac{a^3+b^3+c^3}{abc}=6$$ holds, then also

$$\frac{h^3a^3+h^3b^3+h^3c^3}{hahbhc}=6$$ does. In fact you can group $$h^3$$ both above and beneath and then simplify.

One last thing to mention. Of course, while this does not help us in finding the three base numbers, it tells us that, once found, the number a+b+c=n has a perfect cube. Moreover, each number m=hn, with $$h \in \mathbb N$$ is a perfect cube either.

That's it. I don't know where this has been proven, because this is only the result of my spare time calculations. I hope some of you might find it interesting enough to share opinions and ideas. Now I'm trying the same for the more general rule:

$$\sum_{i=1}^nx_i^n = y^n$$

but it already proved to be nasty!

Answer 3

This is more of a comment as opposed to an answer

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There is a formula for finding the values of $a, b, c, d$ in the following equation: $$a^3 + b^3 + c^3 = d^3$$ Where $$\forall x, y\in \mathbb{Z}, \ \begin{align} a &= 3x^2 + 5y(x - y), \ b = 2\big(2x(x - y) + 3y^2\big) \\ c &= 5x(x - y) - 3y^2, \ d = 2\big(3x^2 - 2y(x + y)\big) \end{align}$$

Therefore if we prove this conjecture, we prove that every perfect number $d$ must be even! I also would like to expand this theorem as well with a theorem of mine:

If $$\forall\{x, y, z\}\subset \mathbb{N}, \ 6^3 + (2x)^3 + (2y - 1)^3 = z^3$$ Then $$z \equiv \pm 1 \pmod 6$$ Where $z$ is a prime number. If $z$ is not a prime number, then $z\equiv3\pmod 6$

Examples: $$\begin{align} 6^3 + 8^3 + 1^3 &= 9^3 \qquad \ \ \ \ \mid9 &\equiv 3 \pmod 6 \\ 6^3 + 32^3 + 33^3 &= 41^3 \qquad \ \ \mid41 &\equiv 5 \pmod 6 \\ 6^3 + 180^3 + 127^3 &= 199^3 \qquad \mid199 &\equiv 1 \pmod 6 \\ 6^3 + 216^3 + 179^3 &= 251^3 \qquad \mid251 &\equiv 5 \pmod 6 \\ 6^3 + 718^3 + 479^3 &= 783^3 \qquad \mid783 &\equiv 3 \pmod 6 \\ 6^3 + 768^3 + 121^3 &= 769^3 \qquad \mid769 &\equiv 1 \pmod 6 \end{align}$$

Answer 4

It can be shown that if $\sigma$ is an even perfect number then $\sigma^3$ can be expressed as a sum of three positive cubes. Since there are no known odd perfect numbers, this also means that the cubes of all known perfect numbers can be so expressed. But it doesn't guarantee that every perfect number can be so expressed, since it hasn't been proved that there are no odd perfect numbers (although that seems very likely - see here).

To prove the proposition for the even case, first note that (as pointed out in Thomas Andrews' comment) an even perfect number $\sigma$ must be of the form $2^{p-1}(2^p-1)$ where $p$ and $2^p-1$ are prime (see Mathworld, equation (15) and following paragraph). Hence any $\sigma$ with $p>8$ has $2^8$ as a factor. But:

$$(2^8)^3=256^3=9^3+183^3+220^3$$

Hence for any such $\sigma$:

$$\sigma^3= (2^82^{p-9}(2^p-1))^3=(9(2^{p-9})(2^p-1))^3+(183(2^{p-9})(2^p-1))^3+(220(2^{p-9})(2^p-1))^3$$

It remains to show that the proposition holds when $p\leq8$. But the only such cases are $p=2,3,5,7$ yielding respectively $\sigma=6,28,496,8128$ which have already been dealt with in the question and, for $8128$, in the answer by Tito Piezas III.