# Talk:Erdős–Straus conjecture

The Swett link appears to be broken. — Preceding unsigned comment added by 63.229.7.84 (talk • contribs) 02:41, 20 September 2005 (UTC)

I very much doubt the correctness of this section in Generalization: Hagedorn proved a related conjecture of R. H. Hardin and Neil Sloane that, for odd positive n, the equation 3/n = 1/x + 1/y + 1/z is always solvable with x, y, and z also odd and positive.
A proof of this conjecture would be trivial: Let 
  
    
      
        x
        =
        y
        =
        z
      
    
    {\displaystyle x=y=z}
  
 be 
  
    
      
        n
      
    
    {\displaystyle n}
  
.
Ocolon 22:33, 29 January 2007 (UTC)
x, y, and z must not equal each other. I edited in a clarification. —David Eppstein 03:16, 30 January 2007 (UTC)
I prove this guess is true and if you give me any number I solve it very quickly Mohammad ghanbary (talk) 09:06, 17 January 2021 (UTC)

A link was added to the Leibniz harmonic triangle. I wonder if this is relevant to the topic. The Leibniz harmonic triangle might be used to find solutions for 
  
    
      
        n
        =
        4
        m
      
    
    {\displaystyle n=4m}
  
 but the case is rather simple anyway:

  
    
      
        
          
            4
            n
          
        
        =
        
          
            1
            n
          
        
        +
        
          
            1
            n
          
        
        +
        
          
            1
            
              n
              2
            
          
        
      
    
    {\displaystyle {\frac {4}{n}}={\frac {1}{n}}+{\frac {1}{n}}+{\frac {1}{\frac {n}{2}}}}
  

Ocolon 08:28, 4 March 2007 (UTC)

You stopped with that one case, but it can also be used to get solutions with distinct denominators. And with a little creativity, it can also help when 4|n is false. Anton Mravcek 20:10, 4 March 2007 (UTC)
While this might be true, neither the description of the link nor the article about the Leibniz harmonic triangle says that it can be used to find solutions for natural numbers that are not divisible by 4 or that it provides distinct denominators. Besides, the Erdős–Straus conjecture does not require distinct denominators. Furthermore, I don't understand this article as collection of solutions for special cases of the Erdős–Straus conjecture. Shall we add links that solve it for all Mersenne primes, for all numbers that are divisible by 1234, for all square numbers…
I don't question that the Leibniz harmonic triangle may be used to compute some solutions. I question the relevance of that — and even if it was relevant — the helpfulness of the link as the Leibniz harmonic triangle article stops where I stopped: Naming that simple case. Ocolon 08:32, 5 March 2007 (UTC)
I was wondering the same thing. Also, are there published sources connecting the Leibniz triangle to the 4/n problem, or is it just someone's original research? —David Eppstein 18:28, 4 March 2007 (UTC)
There probably aren't any published sources making the connection. It's too simple. The first Google Scholar result uses a complicated bunch of equations to show why the triangle is symmetrical. Even so, I wouldn't be as dismissive as calling this "just someone's original research." It's probably been rediscovered several times by several different people. Anton Mravcek 20:10, 4 March 2007 (UTC)

The sequence A073101 requires that 
  
    
      
        x
        <
        y
        <
        z
      
    
    {\displaystyle x<y<z}
  
.
In particular, this means that 
  
    
      
        x
      
    
    {\displaystyle x}
  
, 
  
    
      
        y
      
    
    {\displaystyle y}
  
 and 
  
    
      
        z
      
    
    {\displaystyle z}
  
 must be distinct, something which is not required by the Erdös-Straus formulation. --Kuifware (talk) 19:02, 28 March 2008 (UTC)

If there exists a solution 4/n = 1/x + 1/y + 1/z with x,y,z non-distinct, then there exists another solution with them distinct. See e.g. [1]. —David Eppstein (talk) 19:17, 28 March 2008 (UTC)
You are right: for 
  
    
      
        n
        ≥
        3
      
    
    {\displaystyle n\geq 3}
  
 you can construct a solution with distinct denominators from a solution where they are not necessarily distinct.  (For 
  
    
      
        n
        =
        2
      
    
    {\displaystyle n=2}
  
 this is not possible however.)
There is a difference with the proof at your website: for the 
  
    
      
        4
        
          /
        
        n
      
    
    {\displaystyle 4/n}
  
 problem the Egyptian fraction should always consist of exactly 3 terms, so the reduction from 
  
    
      
        1
        
          /
        
        y
        +
        1
        
          /
        
        y
      
    
    {\displaystyle 1/y+1/y}
  
 to 
  
    
      
        2
        
          /
        
        y
      
    
    {\displaystyle 2/y}
  
 for even 
  
    
      
        y
      
    
    {\displaystyle y}
  
 needs to be modified a bit.  For example: write 
  
    
      
        y
        =
        
          2
          
            p
          
        
        
          y
          ′
        
      
    
    {\displaystyle y=2^{p}y'}
  
 with the integer 
  
    
      
        p
        ≥
        0
      
    
    {\displaystyle p\geq 0}
  
 as large as possible (i.e. 
  
    
      
        
          y
          ′
        
      
    
    {\displaystyle y'}
  
 is odd), then the reduction 
  
    
      
        
          
            
              1
              y
            
          
          +
          
            
              1
              y
            
          
          =
          
            
              1
              
                2
                
                  p
                
              
            
          
          ⋅
          
            
              2
              
                y
                ′
              
            
          
          =
          
            
              2
              
                
                  2
                  
                    p
                  
                
                (
                
                  y
                  ′
                
                +
                1
                )
              
            
          
          +
          
            
              2
              
                
                  2
                  
                    p
                  
                
                
                  y
                  ′
                
                (
                
                  y
                  ′
                
                +
                1
                )
              
            
          
        
      
    
    {\displaystyle \textstyle {\frac {1}{y}}+{\frac {1}{y}}={\frac {1}{2^{p}}}\cdot {\frac {2}{y'}}={\frac {2}{2^{p}(y'+1)}}+{\frac {2}{2^{p}y'(y'+1)}}}
  
 can be used if 
  
    
      
        
          y
          ′
        
        >
        1
      
    
    {\displaystyle y'>1}
  
, and the reduction 
  
    
      
        
          
            
              1
              y
            
          
          +
          
            
              1
              y
            
          
          =
          
            
              1
              2
            
          
          ⋅
          
            
              1
              
                2
                
                  p
                  −
                  2
                
              
            
          
          =
          
            (
            
              
                
                  1
                  3
                
              
              +
              
                
                  1
                  6
                
              
            
            )
          
          ⋅
          
            
              1
              
                2
                
                  p
                  −
                  2
                
              
            
          
          =
          
            
              1
              
                3
                ⋅
                
                  2
                  
                    p
                    −
                    2
                  
                
              
            
          
          +
          
            
              1
              
                6
                ⋅
                
                  2
                  
                    p
                    −
                    2
                  
                
              
            
          
        
      
    
    {\displaystyle \textstyle {\frac {1}{y}}+{\frac {1}{y}}={\frac {1}{2}}\cdot {\frac {1}{2^{p-2}}}=\left({\frac {1}{3}}+{\frac {1}{6}}\right)\cdot {\frac {1}{2^{p-2}}}={\frac {1}{3\cdot 2^{p-2}}}+{\frac {1}{6\cdot 2^{p-2}}}}
  
 can be used if 
  
    
      
        
          y
          ′
        
        =
        1
      
    
    {\displaystyle y'=1}
  
 and 
  
    
      
        p
        ≥
        2
      
    
    {\displaystyle p\geq 2}
  
. The remaining two cases, where 
  
    
      
        y
        ∈
        {
        1
        ,
        2
        }
      
    
    {\displaystyle y\in \{1,2\}}
  
, are not covered because there are no Egyptian fractions representations for the numbers 1 and 2 using only two terms.
I found the argument for the finiteness of the reduction procedure interesting, though perhaps not entirely obvious.  But I agree: there can only be a finite number of steps because the steps are lexicographically decreasing (the implication can be proven by induction on the number of terms). --85.144.141.41 (talk) 18:21, 29 March 2008 (UTC)
I slightly modified the text to reflect the fact that A073101 considers distinct denominators.  I did not add a statement pointing out the subtle difference. --Kuifware (talk) 19:13, 29 March 2008 (UTC)
The change looks fine to me. —David Eppstein (talk) 19:46, 29 March 2008 (UTC)

How about a couple of simple examples to illustrate?  — Preceding unsigned comment added by 71.131.188.159 (talk) 03:39, 6 July 2011 (UTC) 
=
Like, for n=4: 4/4 = 1/2 + 1/3 + 1/6
Seems like a good example to get started with  — Preceding unsigned comment added by 71.131.188.159 (talk) 03:42, 6 July 2011 (UTC) 

I don't think 4/4 is a great example because it's not in lowest terms, but I replaced the complicated example in the lead section with a much simpler one, 4/5. Is that close to what you meant? —David Eppstein (talk) 07:27, 6 July 2011 (UTC)

Since nowhere in the article does the word "distinct" appear — and in the initial statement neither "distinct" nor any synonym for it appears — the article does not make clear whether in the equation 4/n = 1/x + 1/y + 1/z the integers x, y, z are required to be distinct or not.
This ought to be obvious to anyone with any experience in writing mathematics, but: It is necessary to disambiguate anything that might readily be misunderstood, like whether x, y, z are required to be distinct. 
For anyone who is certain that they know what this conjecture is: Please clarify explicitly whether  or not x, y, z are required to be distinct in this article — especially right at the top of the article.Daqu (talk) 23:50, 4 December 2014 (UTC)

It makes absolutely no difference whether you require it or don't. The answer is the same either way. If a number has a ≤k-term representation as a sum of unit fractions, it also has a ≤k-term representation as a sum of distinct unit fractions. See e.g. http://www.ics.uci.edu/~eppstein/numth/egypt/conflict.html for why. As for whether we should state this explicitly the first time we formulate the conjecture: my preference would be no, for the same reason that we don't want to add disclaimers stating that x, y, and z are not assumed to be relatively prime, and are not assumed to be perfect numbers, and are not assumed to be in arithmetic progression: there are an infinite number of extra constraints that we could plausibly add, but aren't adding, and we don't have room to write them all down. —David Eppstein (talk) 00:49, 5 December 2014 (UTC)
This needs to be stated explicitly in the article: either a) that x, y, z are not assumed to be distinct, or b) that they are assumed to be distinct. One or the other.  It it really that hard to grasp that clarity is necessary for an encyclopedia article???
Clarity makes every difference to those reading the article.Daqu (talk) 01:20, 6 December 2014 (UTC)
Do we also need to state explicitly that we're not requiring them all to be divisible by 13? If not, what is the reason for stating one non-constraint explicitly and not stating another? If we don't constrain them, they are not constrained. What is confusing about this? And for that matter why do you insist that it be "one or the other" when it doesn't affect the answer? If we add anything more to the article, it should *not* be that they are required to be distinct, nor that they are allowed to be equal, but rather that some authors formulate the problem one way and some formulate it the other and that the distinction is unimportant. Which by the way is already in the article (at the end of the first "background" section paragraph) but maybe not prominently and explicitly enough for your preference. —David Eppstein (talk) 01:25, 6 December 2014 (UTC)
Seriously, do you not recognize how very false your analogies are?  Are you not aware that these articles are not necessarily read by professional mathematicians, and even when they are, they don't expect to find rigorous writing as one might find in a professional journal? Anyone reading the phrase "the sum of three unit fractions" or the phrase "there exist positive integers x, y, and z such that" — both of which appear in the article's initial statement of the conjecture — is bound to wonder one thing: Do x, y, and z need to be distinct?
If the answer to this question is, as you state, that both forms of the conjecture are equivalent, then that should at least be mentioned, so as to dispel confusion..
Further: As regards your important point that: "The answer is the same either way. If a number has a ≤k-term representation as a sum of unit fractions, it also has a ≤k-term representation as a sum of distinct unit fractions.":  Aren't we talking about whether 4/n is equal to the sum of 3 unit fractions, and not merely ≤ 3 unit fractions?
Only a few seconds are lost in writing a clarifying sentence. Everything is lost if future readers are left unsure of what the conjecture is.Daqu (talk) 17:42, 6 December 2014 (UTC)
A few seconds THAT HAVE ALREADY BEEN TAKEN. Or, what don't you like about the clarifying sentence that is already there, at the end of the first paragraph of the background section? And frankly I find your insistence that beginning mathematics readers will be confused by this and that this will be an earthshaking calamity to be overblown and odd. Nowhere else in mathematics do we ever assume that variables are disallowed from having the same value without saying so explicitly. If readers come into this article knowing what a variable is but not knowing that, they have much bigger problems than failure to understand a number-theoretic conjecture. The extra constraint that the values be distinct doesn't come from algebra, it comes from the history of the problem and its connection to Egyptian mathematics. So the background section describing that connection is exactly where to treat this material. —David Eppstein (talk) 18:22, 6 December 2014 (UTC)

In the section Negative-number solutions it says

The restriction that x, y, and z be positive is essential to the difficulty of the problem, for if negative values were allowed the problem could be solved trivially via one of the two identities

  
    
      
        
          
            4
            
              4
              k
              +
              1
            
          
        
        =
        
          
            1
            k
          
        
        −
        
          
            1
            
              k
              (
              4
              k
              +
              1
              )
            
          
        
      
    
    {\displaystyle {\frac {4}{4k+1}}={\frac {1}{k}}-{\frac {1}{k(4k+1)}}}
  

and

  
    
      
        
          
            4
            
              4
              k
              −
              1
            
          
        
        =
        
          
            1
            k
          
        
        +
        
          
            1
            
              k
              (
              4
              k
              −
              1
              )
            
          
        
        .
      
    
    {\displaystyle {\frac {4}{4k-1}}={\frac {1}{k}}+{\frac {1}{k(4k-1)}}.}
  

These have only two terms on the right side, so what are x, y, and z in each case?Loraof (talk) 16:10, 18 February 2016 (UTC)

Making an expansion longer is easy. You could replace one of the two fractions 1/x  in this expansion with either 1/2x + 1/2x or 1/(x + 1) + 1/x(x + 1). —David Eppstein (talk) 17:01, 18 February 2016 (UTC)

It would be nice to give a table or list of numerical examples—each solution for, say, n up to 10 or 15. Loraof (talk) 01:14, 8 July 2017 (UTC)

A table or list of numerical answers even for those solutions up to n = 15 would be too large and would not particularly benefit the article. I get my students to check out the following site about Egyptian Fractions in general (section 6.1.1 is really useful if you are looking for all solutions of the shortest length for a particular value of n): http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fractions/egyptian.html#shortCalc
AirdishStraus (talk) 15:23, 8 July 2017 (UTC)

فرمول این حدس را پیدا کرده ام و این حدس کاملا درست است Mohammad ghanbary (talk) 21:45, 16 January 2021 (UTC)

We use English on talk pages here; see WP:ENGLISHPLEASE. And unless your work has been properly peer-reviewed and published in a respectable mathematics journal, it is original research and off-topic; this talk page is only for discussing article improvements and we can only base our article content on reliable sources. —David Eppstein (talk) 21:54, 16 January 2021 (UTC)
I solve it Mohammad ghanbary (talk) 22:09, 16 January 2021 (UTC)

This review is transcluded from Talk:Erdős–Straus conjecture/GA1. The edit link for this section can be used to add comments to the review.
Reviewer: HenryCrun15 (talk · contribs) 04:27, 8 January 2022 (UTC)


### Volunteering
Hi, I will review this article. HenryCrun15 (talk) 04:27, 8 January 2022 (UTC)


### Review against the GA criteria
Thank you for the opportunity to review this article. I found the subject fascinating. A couple of points before I continue with my review:

My apologies for rushing in and incorrectly editing the article when I first started working on this review. I had clearly skimmed the article and jumped to certain conclusions. Thank you for your corrections
I have a Bachelor's degree in mathematics, but no further, so my ability to comment on highly specialised concepts in number theory will be limited.
I got a Good Article mentor to look over my draft review. Thanks to Nikkimaria (talk · contribs) for the advice, and I have incorporated their suggested changes.
My comments on the article are below:

The article has now been significantly rearranged in reponse to these comments. Some more detailed remarks:

Lead

All comments under 1b done. I also added a sentence at the end summarizing the generalizations section.
Formulation

Fine, both moved. I'm not convinced that moving the polynomial version up is an improvement, but it also doesn't really hurt.
Background

Explanation for why indistinct fractions can be made distinct moved to a new section.
Worst-case length for each numerator and behavior of the greedy algorithm separated into two paragraphs.
Modular identities

Split into two sections and combined with the theoretical part of the "number of solutions" section. The computational part of the "number of solutions" section has instead been merged into the other computation section.
Attempted to explain Hasse principle more clearly
The section on computational verification does not belong in this section, but I did repeat in more detail the explanation for why the search can be limited to primes.
The " one of six values modulo 840" is not a computational result, it is a mathematical result. It is incorrect that this statement appeared only in the lead. It is a summary of a statement later in the article, "except possibly those that are 1, 121, 169, 289, 361, or 529 mod 840".
Negative-number solutions and Generalizations
The negative-number part is not really a generalization. Or if it is it's a trivial generalization that nobody cares about because it's trivial. Really it's an explanation for why the formulation of the main problem uses positive numbers. I moved it into the formulation section. Same goes for the distinct vs not distinct part.
"The proof that the conjecture is true when negative unit fractions are allowed seems incomplete. It shows (two ways) that odd values of n have solutions, but doesn't show how even values have solutions. If so, add this in. ": This was omitted because it's so totally trivial that no source even mentions it. It's just a special case of the expansion for 
  
    
      
        n
        =
        2
      
    
    {\displaystyle n=2}
  
 and the fact that a solution for any 
  
    
      
        n
      
    
    {\displaystyle n}
  
 gives a solution for any multiple of 
  
    
      
        n
      
    
    {\displaystyle n}
  
. No negative numbers needed. (Allowing them doesn't mean they are required to be used.)
Have shown => showed: Done.
Remove "trivially": done.
Add a colon at the end of the sentence.: not done.: We don't end sentences that way.: We also don't put: colons before: nouns in: sentences. The equation in this sentence is, grammatically, a noun. It doesn't need a colon.
External links
Re: "There doesn't seem a need for a link to the Tao blog post in the External links section given it is already linked in the notes.": It is false that it is already linked in the notes. The references link to a technical paper by Tao. The external link goes to a blog post of Tao, describing his work in that paper in a more accessible way (or at least trying to; I don't always find Tao's blog posts to be easy to read). They have the same title but are different links. The notes link to a different blog post by Tao with a different title.
Broadness of coverage
Re "It isn't stated in the article why the Erdős–Straus conjecture is important.": it isn't important, in the sense that (as far as I know) it has no real-world applications. Why would you think that it would? This is not the sort of topic that is studied for its importance. In any case, without sources that make editorial judgements about its importance, we cannot add those judgements to the article.
Re "Why do some researchers investigate this alternative of the conjecture?": It isn't because they want to study a different variation. It's merely that there are two different ways you can define the problem, so when you write down what problem you are defining, you are going to pick one of those two ways. It's an arbitrary choice, but it's not possible to avoid choosing.
Re "If Erdős didn't publish the conjecture until 1950, how was he able to work on the conjecture in 1948? Was he working with Erdős or Straus, or did he hear about it another way?": We can only write what our sources say. Our sources say that it was formulated in 1948 and published in 1950. You know that researchers talk to each other, both formally (in conference talks and seminars) and informally, right? Also that the journal publication process for mathematics papers can take years, with long delays sometimes occurring between coming up with an idea and polishing it to the point that it is suitable for publication, between submitting the paper and getting back a referee report, between that report and the paper's final revision and acceptance, and between acceptance and its actual appearance in the journal? So it is not surprising that Erdős and Straus talked to each other, and maybe to other people like Obláth or maybe to people who talked to other people who talked to Obláth, long before the paper made its way through the long process of getting published. A discussion of the sociology of mathematics publication in the late 1940s is probably well beyond the scope of this article.
Re when did Sierpinski and Schinzel conjecture it: They are both in Sierpinski's 1956 paper. The first part, for 5/n, is stated as "J'oserai aussi poser l'hypothèse que pour les entiers n > 1 tout nombre 5/n est un B3". Here B3 means expressible as a sum of three unit fractions. The phrasing suggests that Sierpinski is making the 5/n conjecture at that point in the paper rather than repeating a conjecture from earlier work. A few lines down he credits the generalization to his student Schinzel: "D'après une hypothèse de A. Schinzel, quel que soit le nombre naturel m donné, il n'y a qu'un nombre fini de nombres naturels n pour lesquels le nombre m/n n'est pas un B3". I added the 1956 date to the article text.
@HenryCrun15: I think now I've addressed all of the comments you made, so could you take another look, please? —David Eppstein (talk) 22:17, 16 January 2022 (UTC)


### Updated review of 17 Jan 2021
Hi @David Eppstein:. I see that you've edited the article following my first review and have requested another look. My updated review is below.

Congratulations on bringing this article to Good Article status!
HenryCrun15 (talk) 05:13, 17 January 2022 (UTC)