Talk:Sums of three cubes

Question about the three very large cubes adding to 3

I put into Google the verbatim expression:

(569936821221962380720)^3 - (569936821113563493509)^3 - (472715493453327032)^3

to see how close to 3 it would calculate it to be.

Instead it returned a value of -4.3735527e+46, rather far from three.

So which one is correct?50.205.142.50 (talk) 02:32, 10 May 2020 (UTC)

The result is exactly 3. You have to do the calculation to enough digits; that is, you need a multiprecision calculator like Pari GP. MathPerson (talk) 02:35, 10 May 2020 (UTC)

Problem with section "Solvability and decidability"

This section refers to a conjecture that each integer ≠ ±4 (mod 9) can be represented in infinitely many ways as the sum of three cubes.

But that section goes on to refer to "the problem" without clearly specifying which problem it is referring to: a) having infinitely many such representations, or b) just having one.50.205.142.50 (talk) 02:37, 10 May 2020 (UTC)

The sentence with the phrase "the problem" in it comes immediately after a sentence that states quite precisely what it means. —David Eppstein (talk) 04:35, 10 May 2020 (UTC)

The "easier" Waring's problem (Wright 1934)

There is an "easier" Waring's problem where the nonnegativity requirement is dropped. It seems relevant both here and there, yet I cannot find it mentioned. Is it mentioned anywhere on Wikipedia? Should it be mentioned in this article? 91.215.142.171 (talk) 18:43, 7 February 2021 (UTC)

I am having some trouble making sense of this question. The article Sums of three cubes (of which this is the talkpage) begins with the sentence ... it is an open problem to characterize the numbers that can be expressed as a sum of three cubes of integers, allowing both positive and negative cubes in the sum, and the body of the article is almost entirely concerned with the problem in which cubes are allowed to be positive or negative. (The positive-only case is considered only in the section Sums of three cubes#Variations, where the connection to Waring's problem is made.) --JBL (talk) 19:04, 7 February 2021 (UTC)

Hello! The "easier Waring's problem" differs from the (usual) Waring's problem in that the summands are allowed to have mixed signs, see e.g. this MO question. I believe it generalizes both the Waring's problem and the sums of three cubes problem. 91.215.142.171 (talk) 19:25, 7 February 2021 (UTC)

Ok, I see, thanks. I presume (particularly because of how you titled this section) that there are published sources about this question. (Unfortunately none seem to be mentioned in that MO thread.) In that case, I think it would be natural to add a section at the end of Waring's problem about it -- to me at least it feels more like a variation of Waring's problem than like a generalization of the three cubes problem. --JBL (talk) 19:37, 7 February 2021 (UTC)

For odd powers such as cubes, there is no difference between the version of the problem where you allow the cubed number to be positive or negative and then only add the cubes (as we do here) and the version where the cubed numbers are positive but then you can add or subtract them (as in the "easier" problem). They are the same problem. So it makes no sense to add another section on a variation that isn't actually a variation. —David Eppstein (talk) 20:31, 7 February 2021 (UTC)

@David Eppstein: I wrote I think it would be natural to add a section at the end of Waring's problem ... and you wrote it makes no sense to add another section .... Am I correct that you mean "to add another section in this article (Sums of three cubes)" (i.e., your comment is not intended as a contradiction of mine)? --JBL (talk) 22:30, 9 February 2021 (UTC)

Oh, I see, I misread it as talking about another section here. Sure, mentioning this in the Waring article makes sense. I assume "easier" refers to the fact that you can represent more numbers this way, not to any evaluation of the mathematical difficulty of studying these problems. —David Eppstein (talk) 22:40, 9 February 2021 (UTC)

"Primitive solutions for n from 1 to 100"

The table only shows solutions up to n=78. Mexicochina (talk) 23:48, 27 April 2021 (UTC)

It also violates WP:DONTHIDE. —David Eppstein (talk) 00:01, 28 April 2021 (UTC)

Some of the solutions aren't primitive (whatever that means). The list - if it should remain in the article - should probably contain the solutions with the smallest cubes. Eg. 2 = 1³ + 1³ + 0³. —ChrisMarch72 (talk) 22:08, 14 October 2022 (UTC)

Please expand

n {\displaystyle n} n cannot equal 4 or 5 modulo 9, because the cubes modulo 9 are 0, 1, and −1, and no three of these numbers can sum to 4 or 5 modulo 9

What cubes?

The cubes. All of them. You know, the numbers 0, 1, 8, 27, etc. —David Eppstein (talk) 04:59, 21 June 2022 (UTC)

cut small example

In this edit I reverted an addition by P0mbal from last year, concerning the equation $3^{3}+4^{3}+5^{3}=6^{3}$. Basically, I feel that the text content (that this equation is "analogous" in some way to the Pythagorean triple $3^{2}+4^{2}=5^{2}$) is misleading (there is no principle here; the "analogy" is just a coincidence), and the connection between that identity and the topic of this article is strained. I would like to see a source that made the connection explicit before restoring it to the article. JBL (talk) 17:57, 4 December 2022 (UTC)

Thank you for paying some attention to this. I dont understand why you metaphorically "threw away the Christams present because you didn't like the wrapping". To get to the point...

(a) Call the identity what you like, include or reject mention of similarities, analogies, etc., however true or apt as you will, the fact is the indentity quoted is true 27+64+125=216 and I think a special case because there isnt one simpler. It seems wrong if those that are aware of it reject the whole idea, rather then editing the verbiage around it. Think what you like about the relation to the Pythagorean triple, a relation which does stand out visibly at least, call it what seems best.

(b) I did not look up that sum of 3 cubes anywhere - it just suddenly occurred to me and astonished me when my 7-y-o grandson was playing with his maths cubes, after we had made the Pythagorean 345 example, and I demonstrated the cube sum with 216 little plastic cubes!

Please understand, and don't throw it away. I could return it an some other form if desired. Thank you. P0mbal (talk) 12:02, 27 December 2022 (UTC)

The Pythagorean triples can be obtained with a formula, not $3^{3}+4^{3}+5^{3}=6^{3}$. And the fact that the value is a cube doesn't have anything special in this article. So I agree with JBL. But for this particular case, perhaps Plato's number could be in the "See also" section. — Vincent Lefèvre (talk) 17:46, 27 December 2022 (UTC)

@P0mbal: I think a special case because there isnt one simpler This article concerns the following question: which integers can be written as a sum of three cubes of integers (positive or negative)? The formula $3^{3}+4^{3}+5^{3}=6^{3}$ is the simplest example of something else (maybe of several other things), but it is not the (or even "a") simplest example of the phenomenon discussed in this article.
I've included in the see-also section a link to Euler's sum of powers conjecture#k = 3, where $3^{3}+4^{3}+5^{3}=6^{3}$ appears as the very first line. I would be perfectly happy to also include the link Vincent Lefèvre suggests in the see-also section. --JBL (talk) 17:20, 28 December 2022 (UTC)

Thank you both. But doesnt the formula look good and jump out at you a something pretty astonishing. The sum of these three successive small cubes making another cube, isnt that worth a mention is this particular article, with the sum before your eyes, even without comment? I know that there are other sets of 3 integer cubes that make another integer cube, but is there another as neat, say as notable as this? That with links to Euler's sum of powers conjecture, Plato's number and the Pythagorean triples would be good. Thank you. — Preceding unsigned comment added by P0mbal (talk • contribs)

I agree with you that it is an interesting, attractive, and surprising formula. So is $e^{i\pi }+1=0$ and obviously that doesn't belong in this article. The question of writing an integer cube as the sum of three cubes is, from the point of view of this article, trivial: we always have $a^{3}=a^{3}+b^{3}+(-b)^{3}$ for every integer b. The Pythagorean Theorem really has nothing at all to do with this article except that both involve adding powers of integers. I'll add the other link in See Also. --JBL (talk) 23:58, 29 December 2022 (UTC)

Definition of n

I spent a long time reading and rereading the first paragraph of this article, because n was used, but not formally defined. After doing a google search and reading what other authors had to say, I began to guess the n meant numbers, and I modified the article to say that. I waited to see if I would be reverted for making a mistake. Instead @JayBeeEll reverted me, because I guessed correctly! Thanks to @JayBeeEll, I now understand this article. How can we fix this article so that other readers do not struggle the way I did? Comfr (talk) 18:43, 5 August 2023 (UTC)

Here is my proposal:

In the mathematics of sums of powers, sums of three cubes is an open problem to determine if a given integer $n$ can be expressed as a sum of three cubes of integers, allowing both positive and negative cubes in the sum. Comfr (talk) 00:21, 6 August 2023 (UTC)

Hi Comfr, I see your point; let me try something and maybe it will help? --JBL (talk) 17:45, 6 August 2023 (UTC)

Also, Comfr, see MOS:BOLDAVOID: "If the article's title does not lend itself to being used easily and naturally in the first sentence, the wording should not be distorted in an effort to include it." —David Eppstein (talk) 07:07, 7 August 2023 (UTC)

I only replaced the pronoun "it" with the antecedent. I did not distort anything. The lede is already quite confusing and distorted. I had a very hard time trying to understand it. Comfr (talk) 08:38, 7 August 2023 (UTC)

"Sums of three cubes" is not the antecedent of "it" in the first sentence. I certainly believe that the lead could be made clearer. You did a very clear job identifying your problem with the introduction of n (and I hope you find it clearer now with my adjustment); perhaps you could try to articulate what else about it you find challenging? --JBL (talk) 17:35, 7 August 2023 (UTC)

In fact there is no antecedent of "it". Instead, "it" has a postcedent, "to determine if...". The sentence could be rewritten in a more direct form as something like "To determine if a given integer can be expressed as a sum of three cubes of integers, allowing both positive and negative cubes in the sum, is an open problem". Using "it is" allows the simpler and more important part of the sentence, "an open problem", to come first, saving the technicalities of positive and negative cubes for the end of the sentence where they can be more easily skipped over. That is, this grammar is intended to make the sentence easier to read. —David Eppstein (talk) 21:52, 7 August 2023 (UTC)

You are correct. At last, I understand the semantics. The word "it" is used as a dummy pronoun, rather than a referential pronoun. The difficulty occurs is because the title of the article and the subject of the article are not the same. The title is "Sums of three cubes", but the subject is "The mathematical problem commonly called, Sums of three cubes". With research papers the title and subject are usually the same for example, "Sum-of-Three-Cubes Problem Solved for ‘Stubborn’ Number 33". Comfr (talk) 22:27, 7 August 2023 (UTC)

The title of the article and the subject of the article are the same. The subject is sums of three cubes. That is, to spell it out in more sesquipedalian, lengthy, and non-title-worthy terms, the integers that can be expressed as a sum of the cubes of three integers. The numbers ..., -8, -7, -6, -3, -2, -1, 0, 1, 2, 3, 6, 7, 8, ... . The unsolved problem of proving that these are the same as the numbers that are ≠ 4,5 mod 9 is a subtopic, not the main topic. But it would be fatuous to say in the lead "A sum of three cubes is an integer that can be expressed as the sum of the cubes of three other integers". Writing it that way would violate a different and nearby part of our Manual of Style on lead sentences, MOS:REDUNDANCY. —David Eppstein (talk) 23:03, 7 August 2023 (UTC)

I see your point. Unfortunately, I could not understand what the article was about until after consulting external sources. External sources should be used to get more details, not to understand what the article is about. Now that I understand the subject matter better, I can see why I got tripped up by the way the article is worded. Thanks for reviewing my attempts to fix the article. The accuracy Wikipedia depends upon editors like you. Comfr (talk) 01:02, 8 August 2023 (UTC)

Typo in solution for 906

The solution for 906 has a typo in it: there are three extra digits in the second number, which should be 72,054,089,679,353,378, see the solution published in https://www.pnas.org/doi/10.1073/pnas.2022377118. I cannot edit the page to fix it, but possibly someone with more privileges can. — Preceding unsigned comment added by 104.129.138.112 (talk) 12:44, 29 May 2024 (UTC)

Semi-protected edit request on 23 May 2024

correct 72 054 821 089 679 353 378 to 72 054 089 679 353 378 Retrowikixvii (talk) 04:31, 3 June 2024 (UTC)

 Done Liu1126 (talk) 06:04, 3 June 2024 (UTC)

I have found new ones

I have found these infinite family of solutions

For all Integers, $a$

$(a^{3}+6x^{3})^{3}+(a^{3}-6x^{3})^{3}+(-6ax^{2})^{3}=2a^{9}$

$(4a^{3}+3x^{3})^{3}+(4a^{3}-3x^{3})^{3}+(-6ax^{2})^{3}=128a^{9}$

$(9a^{3}+2x^{3})^{3}+(9a^{3}-2x^{3})^{3}+(-6ax^{2})^{3}=1458a^{9}$

$(36a^{3}+x^{3})^{3}+(36a^{3}-x^{3})^{3}+(-6ax^{2})^{3}=93312a^{9}$

Could someone verify them and add it ? Potato Imaginator (talk) 02:36, 14 August 2024 (UTC)

You could check using Wolfram Alpha Potato Imaginator (talk) 02:39, 14 August 2024 (UTC)

@Potato Imaginator: Wikipedia does not publish original research. It has to appear in some other reliable publication first. Bubba73 ^{You talkin' to me?} 03:32, 14 August 2024 (UTC)

In any case, this would not be particularly interesting. — Vincent Lefèvre (talk) 08:07, 14 August 2024 (UTC)

I'll publish this paper, and Wikipedia page says "However, 1 and 2 are the only numbers with representations that can be parameterized by quartic polynomials as above." , which is wrong. Potato Imaginator (talk) 08:34, 14 August 2024 (UTC)

@Potato Imaginator:. Yes, but it seems that something is missing in this sentence. Your solutions seem to be particular cases of the "obvious" solutions given at the beginning of Mordell's paper (Eq. (6) and (7): $x=a+bt^{3}$, $y=a-bt^{3}$, $z=-ct^{2}$ where $6ab^{2}=c^{3}$). I have access only to the first page of this paper. — Vincent Lefèvre (talk) 16:40, 14 August 2024 (UTC)

As our article says, "These [representations of 1 and 2] can be scaled to obtain representations for any cube or any number that is twice a cube". --JBL (talk) 17:13, 14 August 2024 (UTC)

However, the above identities do not correspond to a scaling (though the result is still a cube or twice a cube). — Vincent Lefèvre (talk) 17:29, 14 August 2024 (UTC)

Here is the precise theorem statement from the reference:

The equation $x^{3}+y^{3}+z^{3}=n$ has no solutions with $x,y,z$ as quartic polynomials in a parameter $t$ with rational coefficients unless $n=2a^{3},$ or $n=a^{3},$ where $a$ is a rational number, and then the only solutions are those given essentially by [four equation references including (6) and (7) mentioned by VL].

--JBL (talk) 17:38, 14 August 2024 (UTC)

Ah, yes, you're right. They're particular cases of Mordell's paper. Didn't realize that :) How come the general formula from Mordell's paper isn't mentioned here? Potato Imaginator (talk) 02:46, 15 August 2024 (UTC)

@Potato Imaginator and JayBeeEll: Yes, and there are also presentation issues, so this needs some rewrite. But I've just noticed that the theorem given by JBL is incomplete: in "the only solutions are those given essentially by [four equation references including (6) and (7) mentioned by VL]", (6) and (7) are those only for $n=2a^{3}$. Those for $n=a^{3}$ are missing. Are they only those obtained by the formula given in the article, scaled by a cube, or are there other families of solutions? Basically, the other two equation references are missing. — Vincent Lefèvre (talk) 10:35, 15 August 2024 (UTC)

Apologies, it was pure laziness on my part to not re-copy the families in Mordell's theorem. They are

$${\begin{array}{llll}x=a+bt^{3},&y=a-bt^{3},&z=-ct^{2},&{\text{where }}6ab^{2}=c^{3};\\x=t,&y=-t,&z=a;\\x=9at^{4},&y=3at-9at^{4},&z=a-9at^{3}.\end{array}}$$

(Not sure why Wiki-LaTeX is jamming in a bunch of extra colons, sorry.) The first line is the family we've been discussing (equations (6) and (7)). The next two lines give the two families of solutions for $a^{3}$. (Of course different-looking solutions are possible if we make a substitution $t\mapsto dt+e$.) --JBL (talk) 19:30, 15 August 2024 (UTC)

You shouldn't put the colons (used for indenting in wiki code only) inside <math>...</math> tags, where everything is interpreted as LaTeX (not wiki code). I've removed these colons. — Vincent Lefèvre (talk) 19:43, 15 August 2024 (UTC)

Fascinating -- because the "reply tool" didn't understand that those lines of text were part of a math code block, it tried to indent them as normal text. Shows what I get for not using the source editor. Thanks for fixing! --JBL (talk) 19:49, 15 August 2024 (UTC)