17
$\begingroup$

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?

$\endgroup$
20
  • 1
    $\begingroup$ Where did you find this conjecture? Or did you just notice this for the first few cases? $\endgroup$ – Mark Fischler Jan 25 '15 at 4:15
  • 5
    $\begingroup$ None of $495$, $82$, and $57$ are prime. $\endgroup$ – Micah Jan 25 '15 at 4:22
  • $\begingroup$ See artofproblemsolving.com/Forum/… $\endgroup$ – AsdrubalBeltran Jan 25 '15 at 4:29
  • 1
    $\begingroup$ How far have you checked this? Note that here oeis.org/A023042 it looks like a large proportion of cubes are the sum of three cubes. $\endgroup$ – Eric M. Schmidt Jan 25 '15 at 4:31
  • 1
    $\begingroup$ The even powers of $2^{2k}$ are congruent to $1$, $4$, $7$, $1$, $4$, and so on mod $9$. The corresponding $2^{2k+1}-1$ are congruent to $1$, $7$, $4$, $1$, and so on. The product is congruent to $1$, $1$, $1$, and so on. There is a famous conjecture that all numbers not of the form $9k\pm 4$ are a sum of three cubes. That would take care of even perfects. $\endgroup$ – André Nicolas Jan 25 '15 at 5:05
15
$\begingroup$

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 $99\text{%}$.

$\endgroup$
13
  • $\begingroup$ So what I said is correct but can the proof that all these $\endgroup$ – user210387 Jan 25 '15 at 16:42
  • $\begingroup$ Perfect numbers can be infinitely expressed as sum of three cubes is it present or do I have to prove it $\endgroup$ – user210387 Jan 25 '15 at 16:43
  • $\begingroup$ we find that the cube every perfect number can be expressed as sum of cubes in infinitely many ways not because its perfect but because as you say the odds are in favour but we cannot find a general proof that these cube of perfect numbers can be expressed as the sum of three cubes in infinitely many ways. So my question is still there because I want a general proof for thoa $\endgroup$ – user210387 Jan 25 '15 at 16:52
  • $\begingroup$ @SayanChattopadhyay: I said, "...chances are it can be expressed as a sum of positive cubes in many ways as well." Where did you get the infinitely many part? Surely you see there can't be infinitely many positive integers $a,b,c$ such that $a^3+b^3+c^3$ is a constant sum? (And, yes, it is not a water-tight proof for perfect numbers. There might not any at all, since the sum of three positive cubes property is due more to statistics, rather than on its "perfectness".) $\endgroup$ – Tito Piezas III Jan 25 '15 at 23:01
  • $\begingroup$ So there will be no proof at all $\endgroup$ – user210387 Jan 26 '15 at 3:02
2
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

This is more of a comment as opposed to an answer


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

$\endgroup$
1
  • $\begingroup$ How did you derive your formulae for $a$, $b$, $c$, and $d$? Were you able to prove that it is the only such formula, @MrPie? $\endgroup$ – Arnie Bebita-Dris Jan 22 '20 at 23:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy