Talk:Abc conjecture/Archive 1

Fermat
I've read that Fermat's Last Theorem is an easy consequence of this conjecture. If I get the chance I'll see if I can find out why and add it, unless someone else wants to do so in the meantime. --ScottAlanHill 16:54, 9 March 2006 (UTC)

Since the conjecture says nothing on the constant $$C_{\epsilon}$$, I don't see how it should rule out the existence of finitely many solutions for FLT. However, the abc-conjecture does easily imply that for n>3 there are at most finitely many solutions: If X,Y,Z are positive integers without common divisor with $$X^N+Y^N=Z^N$$, then $$rad (X^NY^NZ^N)=rad(XYZ)$$ and $$rad(XYZ) \le Z^3$$ and therefore the abc-conjecture (if true) yields
 * $$ Z^N < C_{\varepsilon} \operatorname{rad}(X^NY^NZ^N)^{1+\epsilon}

\le C_{\varepsilon} Z^{3+3\epsilon}, $$ which implies an upper bound for Z.

Joerg Winkelmann 20:22, 12 May 2006 (UTC)


 * Your comment is not relevant to Wikipedia, unless it's been published in a peer-reviewed journal that can be referenced as a reliable source. -- 184.189.217.210 (talk) 08:15, 19 November 2017 (UTC)
 * Joerg, Ribenboim gives the same calculation in Fermat's Last Theorem for Amateurs, chapter XI, (he uses $$ \epsilon = 1/2 $$), but the last step is different: the last eqn. gives an upper bound for $$N$$ (not $$Z$$), viz.
 * $$ N < ln(C_{\varepsilon})/ln(Z)+3+3\epsilon < ln(C_{\varepsilon})/ln(2)+3+3\epsilon = N_0 $$
 * where I used $$ Z > 2 $$, but a much higher minimal $$Z $$ should be possible. Ribenboim: ...in other words, FLT is true for every exponent $$ n>n_0 $$, or in short, FLT is asymptotically true.
 * And in case anyone is interested, the comment of IP-man 184.189.217.210 is quite daft. WP should block IP-users, they only make a mess. Herbmuell (talk) 17:58, 14 April 2022 (UTC)

Math-mode formulas
Is there a reason not to use math formulas here? These are the math-style formulas: Vegard 14:01, 29 September 2006 (UTC)
 * a + b = c
 * $$a + b = c$$
 * rad(abc),
 * $$\operatorname{rad}\,(abc)$$
 * rad(abc)/c
 * $$\frac{\operatorname{rad}\,(abc)}{c}$$
 * rad(abc)1+ε/c
 * $$\frac{{\operatorname{rad}\,(abc)}^{1+\epsilon}}{c}$$

a question
if a, b, c are positive integers, why take the absolute value |a| + |b| + |c| ?


 * Good point. I think this is a better version of that section, but I don't actually know the conjecture. If this is good, maybe someone can put it in the article:

Formulation
For any integer n, the radical of n is defined to be the square-free product of its distinct prime factors. In other words, the product of all of the unique prime factors of n, never raising a factor to a power greater than 1. It is denoted rad(n).

In the article the definition of the conjecture has the words "For every ε > 0" I think this is misleading, The words should be "for any ε > 0". That way it is clear why "if the conjecture is true then there must exist a triple (a, b, c) which achieves the maximal possible quality q(a, b, c)". — Preceding unsigned comment added by 2A02:ED0:5FF4:100:4D9A:16B7:A59E:44AA (talk) 23:01, 12 September 2018 (UTC) The abc conjecture states that, for any ε > 0, there exists a finite Kε such that, for all coprime positive integers a, b and c such that a + b = c,


 * $$a+b+c<K_\epsilon\operatorname{rad}(abc)^{1+\epsilon}.$$

Orthografer 17:59, 8 May 2007 (UTC)

Solution
Lucien Szpiro announced a solution to this at the Goldfeld conference. 146.96.245.119 20:42, 24 May 2007 (UTC)

Unclear Wording
In section Some consequences, the term first group in the last sentence seems unclear to me.
 * While the first group of these have now been proven ...'

-- Burkhard.Plache 15:48, 6 September 2007 (UTC)

Granville-Langevin conjecture
"Granville-Langevin conjecture" now redirects here, and should be explained here or forked into its own article. -- Beland 21:06, 1 December 2007 (UTC)


 * This reference may help: . Thue Siegel Roth (talk) 15:50, 14 June 2013 (UTC)


 * Added as reference  Spectral sequence (talk) 17:23, 14 June 2013 (UTC)

Dabrowski example
The Dabrowski example seems less significant than the others. Richard Pinch (talk) 21:51, 13 October 2008 (UTC)

Partial results
Are we sure about those? I mean, there is no "exp" in the conjecture, though the partial results are...well lets say much weaker. How comes they are called "partial results"? Exponential instead of linear... Do I miss something? --Xario (talk) 03:13, 27 October 2008 (UTC)

The word "coprime" is unclear when applied to three or more integers
The word "coprime" is used numerous places in the article, and in particular in defining the ABC Conjecture. (Which, by the way, it is mindlessly idiotic to call the "Abc conjecture", Wikipedia style or not.)

But to call three integers "coprime" is ambiguous.

Does it mean

a) There exists no prime that divides all of the integers?

or

b) For each pair among the integers, there is no prime that divides each of that pair?

(Of course the same comment would apply to any number >= 3 of integers.)

The word "coprime" in the article is linked to the article Coprime, but this deals only with pairs of integers, so is of no help in clarifying its meaning when applied to three integers.Daqu (talk) 21:52, 24 November 2009 (UTC)


 * The first occurrence of coprime in abc conjecture has a note linking to abc conjecture which explains that it doesn't matter which meaning of coprime is used in the conjecture, because a + b = c. The capitalization "Abc conjecture" is not used anywhere in abc conjecture, but technical limitations in the MediaWiki software means that it is used in some other places with automatically created content, for example categories. I have added lowercase title to this talk page to make it change "Talk:Abc conjecture" to "Talk:abc conjecture" at the top, like the article already does. PrimeHunter (talk) 23:33, 24 November 2009 (UTC)


 * Thank you for pointing out to me that whichever definition of coprime is used, the conjecture is equivalent.


 * Still, that's no excuse for using ambiguous language. Disambiguation should be done in the text, not in a footnote.  Footnotes are for parenthetical comments.  One should be able to read a sentence and know what it means because the sentence is clearly written.


 * Because the very word "coprime" is ambiguous when applied to more than two integers at a time, it shouldn't be. Rather, what is meant should just be spelled out.Daqu (talk) 10:35, 26 November 2009 (UTC)

Highest quality triples
This list is misnamed. It should be called "Unbeaten triples". It contains those triples (a,b,c) such that there is no larger triple with higher quality. It does not show the highest quality triples in descending order. For example, there is a triple with c=7*5^4 which has a higher quality than the triple Nr. 4 (c=2*3^3*​5^23​*953) in the list - but c=7^5^4 gets beaten by a larger triple where as c=2*3^3*​5^23​*953 is unbeaten.

So either this list should be renamed ("unbeaten triples" with definition of the term unbeaten) or the highest quality triples as in http://www.math.leidenuniv.nl/~desmit/abc/index.php?set=2 should be taken.--Thn2010 (talk) 12:37, 21 February 2012 (UTC)

Could the note after the quality table be changed from Note: the quality q(a, b, c) of the triple (a, b, c) is defined above. to Note: the quality q(a, b, c) = log(c) / log(rad(abc)) of the triple (a, b, c) is defined above. so the reader doesn't have to scroll up to remind themselves of the definition of q?Kdpw (talk) 11:19, 10 October 2021 (UTC)

Thanks for including these interesting tables. There are no cases listed among the better qualities where both summands are of high potency. Are these cases all sufficiently well behaved or what is the reason? Or is that even the essence of the abc-conjecture? - In the second table, I miss a column that would give the number total of possible representations of c as a+b (a,b coprime). It would of course be interesting to see how the number of quality-triples varies not only with c, but also with the number of possible decompositions of c. As you've counted the qualities, these numbers should be available to you. But probably, there's a formula. Generisches Maskulinum (talk) 13:06, 3 December 2023 (UTC)

On Dabrowski's result
In 2002 Florian Luca published the article "The Diophantine Equation P(x) = n! and a Result of M. Overholt" proving that abc conjecture implies that all diophantine equations in the form P(x) = n!, where P is a polynomial with integer coefficients, deg P > 1, have only finitely many integer solutions. Both Dabrowski's and Luca's work use the approach developed by M. Overholt, so it seems strange to quote Dabrowski's result (which is just the special case when P(x) = x^2 + A) and not Luca's. --95.78.194.243 (talk) 11:57, 27 July 2012 (UTC)


 * Feel free to correct the mistake. Please, be bold! --bender235 (talk) 14:49, 27 July 2012 (UTC)

Heads up
There is a preprint claiming to prove abc by Shin Mochizuki at http://www.kurims.kyoto-u.ac.jp/~motizuki/Inter-universal%20Teichmuller%20Theory%20IV.pdf, discussed by Jordan Ellenberg, Brian Conrad, Terence Tao, Noam Elkies, and others at http://quomodocumque.wordpress.com/2012/09/03/mochizuki-on-abc/ and by Timothy Gowers at https://plus.google.com/103703080789076472131/posts/j1sEGnPyiRu — I don't think we should add anything to our article until either it is peer reviewed and published or the splash gets too big to ignore, but this seems to be a serious attempt. —David Eppstein (talk) 15:18, 4 September 2012 (UTC)


 * I agree that this should only be added if it either received significant news coverage or has been published in a peer-reviewed journal. So lets wait and hope he did it.... -- Toshio Yamaguchi (tlk−ctb) 12:23, 5 September 2012 (UTC)


 * I agree, let's wait. Remember the Deolalikar P/NP proof of two years ago. --bender235 (talk) 15:12, 5 September 2012 (UTC)
 * MO thread. I don't think this situation is similar to Deolalikar, but yeah, it's best to wait til knowledgeable theorists have something to say.  I'm not too fussy about formal publication as long as the info is coming from real experts. 67.119.15.30 (talk) 03:14, 8 September 2012 (UTC)
 * If there is a consensus here, should we remove the mention of Mochizuki's preprint from the lede for the time being?--Ymblanter (talk) 12:26, 16 September 2012 (UTC)
 * I agreed earlier and still do. Its too early to mention this so prominently. --bender235 (talk) 12:36, 16 September 2012 (UTC)
 * Since I made my earlier comments, this has been written about very prominently in Science and Nature. I think that's too big to be ignored any more. —David Eppstein (talk) 16:28, 16 September 2012 (UTC)

Dots as thousands separators in table?
Since this is en.wikipedia.org, shouldn't they be commas? Regardless of the table being quoted from Dutch? 68.44.132.25 (talk) 02:26, 11 September 2012 (UTC)
 * Fixed. Owen&times; &#9742;  04:41, 11 September 2012 (UTC)

not understandable
It is really nice for the mathematicians out there that this article is mathematically correct etc., but for those without that solid mathematical background who hear about this ABC Conjecture solution and want to know what is about, the current description is totally incomprehensible. No, I cannot change it myself because I have no idea what it is all about. — Preceding unsigned comment added by 212.159.206.52 (talk) 07:52, 11 September 2012 (UTC)
 * The lede employs nothing beyond junior-high-school-level arithmetic. Which part exactly confuses you? Owen&times; &#9742;  13:12, 11 September 2012 (UTC)
 * I think the second sentence of the lede expresses the conjecture clearly and concisely without using jargon. The infobox expressing the conjecture was titled "list of unsolved problems in mathematics", and this was confusing.  I moved the link to unsolved problems to a new further reading section and put the conjecture itself under "formulations".  I also added an example of numbers a,b,c, and d to show what they mean and how they are employed.  In my example, c is not greater than d(1+ε).  The conjecture actually states that this is the case most of the time, but an example where c>d(1+ε) might be more useful.  Anyway, a reader should be able to get a good idea of what the conjecture is by reading 7 lines into the body.--Wikimedes (talk) 23:12, 22 September 2012 (UTC)
 * Not quite. I studied math for three years at a highly respectable European university and never came across the term coprime [which, by the way, is red underlined in this Wikipedia write program].2600:6C67:1C00:5F7E:6DF1:CD52:7E5F:D359 (talk) 15:33, 9 July 2022 (UTC)

Clarification of Dabrowski result
In abc conjecture, the result of Dabrowski has been tagged as needing clarification. I don't have the cited paper, but my guess is that n is an positive integer (since the factorial is commonly applied to integers) and A is stated to be an integer, thus k is also an integer. If this is correct, then I think this should perhaps be stated explicitly. Can someone verify that this is correct? -- Toshio Yamaguchi (tlk−ctb) 14:54, 9 October 2012 (UTC)
 * The Zentralblatt review begins For positive integers x,y,A (not a square), the diophantine equation in the title has at most finitely many solutions.  Deltahedron (talk) 17:00, 9 October 2012 (UTC)

Inter-universal geometry? or IU Techmüller theory?
Various media and blogs, e.g. New Scientist & MathOverflow, suggested that he (Shinichi Mochizuki)† used “Inter-universal geometry” (on which we have no article as of 2012-10-27), others that his papers on “Inter-universal Teichmüller theory” are where he proves it (or claims to)†. Is this a mix-up on someone’s part? He does have papers on both, but IUG is a handwritten PDF of what look like lecture notes, while those on IUTT are cleanly type-set papers. The fourth of these does indeed include the claim to prove ABC, so that the reference here is correct. What is the difference between the subjects? Some googling on my part found various references, but left me unclear what the relevance of IUG is. Given the references to IUG I mentioned, a disclaimer in this article would be helpful. PJTraill (talk) 19:28, 27 October 2012 (UTC)
 * † 2 clarifications, 1 typo corrected PJTraill (talk) 20:28, 16 January 2024 (UTC)

Proof by Sow?
What to make of this continuous addition by unknown French IP (Sow himself?) and his claimed proof of abc? Someone posted it on MathOverflow, let's see what the experts think of it. --bender235 (talk) 15:36, 25 May 2013 (UTC) Okay, experts think it's spam. --bender235 (talk) 16:26, 25 May 2013 (UTC)

Mochizuki's work
Given the current status of Mochizuki's claims, it seems WP:UNDUE to devote nearly half the introduction -- 140 out of 305 words -- to them. Spectral sequence (talk) 21:09, 14 June 2013 (UTC)

A related conjecture of Andrew Granville
Current text: A related conjecture of Andrew Granville states that on the right-hand side of the equation, we could also put O(rad(abc) Θ(rad(abc))) where Θ(n) is the number of integers up to n divisible only by primes dividing n. The only equation here is a+b=c. Presumably the upper bound on c is meant? What is the reliable source that would resolve this? Spectral sequence (talk) 16:56, 8 July 2013 (UTC)
 * Fixed now, from Baker (1998). Spectral sequence (talk) 18:23, 8 July 2013 (UTC)
 * How does the line with the Omega notation constitute an upper bound on c? Omega is usually used for lower bounds. Is there a missing assumption that K < 1? —David Eppstein (talk) 19:56, 8 July 2013 (UTC)
 * The very next line reads where Ω(n) is the total number of prime factors of n. Spectral sequence (talk) 20:34, 8 July 2013 (UTC)
 * Ok, it's very confusing to me to see Omega and Theta, used in the context of upper bounds that could well be asymptotic, not meaning Omega or Theta notation, at the same time as using O for O notation — is there another letter we can use or are we stuck with this notation? Also, what is K? Is it less than or greater than one? Is it a constant whose value is known, or just known to exist? —David Eppstein (talk) 21:19, 8 July 2013 (UTC)
 * Omega is quite standard for number of primes counted with multiplicity: Theta is ad hoc, I think. Both are used by the reference cited and both are explained in the article on the line after they are used.  K is a conjectural positive absolute constant (lambda in the reference).  Spectral sequence (talk) 21:27, 8 July 2013 (UTC)

Related conjecture of Robert Stewart Tenenbaum
Current text: I added the most refined version of the conjecture as of 2014. To my embarrassment, it is my first time editing in Wikipedia and I don't know the proper citation. I added a reference link and their result, but the hyperlinks seems not working. If anyone knows how to link properly, it would be much appreciated. You may remove this part as it is done.

I'm wondering if it would be proper to remove the previous conjectures, as those were covered by this new result. — Preceding unsigned comment added by 118.33.68.38 (talk) 07:23, 1 August 2014 (UTC)


 * No, the older material should stand. At the very least it shows the historical and logical development of the topic, and unless and until it is the consensus of academic sources that a new formulation is indeed the most useful, as an encyclopaedia we report what is currently the mainstream view of the subject.  This new version has not yet even been formally published.  Deltahedron (talk) 07:14, 3 August 2014 (UTC)

Quote: This equation a^n + b^n = c^n is the Fermat's Last theorem.
I don't think so. First of all, Fermat's Last theorem is no equation. And if it was something like an equation (say an inequality) then it would be $$a^n+b^n \ne c^n$$. --Jobu0101 (talk) 11:17, 14 February 2015 (UTC)
 * I agree that the sentence is poorly worded, but I'm not sure why it wouldn't qualify as an equation though. It's an equation with no integer solutions for n greater than 2, but an equation nonetheless. Really though, if you can think of a better wording, just go for it. If you feel the edit needs an explanation, go ahead and do so in the edit summary or here. - Tga (talk) 05:19, 18 February 2015 (UTC)

FLT states that there is no n>2 such that x^n+y^n=z^n for positive integers x,y,z,n. Hence it is not an equation, but an ontological statement about an equation. — Preceding unsigned comment added by Generisches Maskulinum (talk • contribs) 11:05, 1 December 2023 (UTC) Generisches Maskulinum (talk) 11:16, 1 December 2023 (UTC)

Example
The example using 1 as a coprime doesn't make sense, in that as I understand number theory, 1 is not prime.98.247.177.223 (talk) 21:59, 18 December 2015 (UTC)
 * 1 is not a prime but 1 is coprime to all integers, i.e. for all integers n, gcd(n, 1) = 1. "coprime" says something about the relationship between two numbers: They have no common prime factor. For example, 4 and 15 are coprime. Being coprime is not a property of an individual number. PrimeHunter (talk) 22:54, 18 December 2015 (UTC)

Equivalence of formulations
The section Formulations gives two formulations "ABC Conjecture" and "ABC Conjecture II" which are said to be equivalent. The second easily follows from the first (if there's only finitely many triples, there must be a best one). But it is far from obvious to me that the first follows from the second. Maproom (talk) 10:07, 26 January 2016 (UTC)


 * I was thinking the exact same thing! Is the equivalence even true? Herpesklaus (talk) 12:37, 28 July 2016 (UTC)

I couldn't agree more. If there are only finitely many cases, then you can of course draw a line somewhere. But that there is a supremum to some set of numbers does not imply that the set is finite. Generisches Maskulinum (talk) 13:28, 3 December 2023 (UTC)

Trivial?
Please add proof that there are infinitely many pairs (a, b) so that a, b and a + b are pairwise coprime. Herpesklaus (talk) 12:00, 28 July 2016 (UTC)
 * It's trivial, and redundant — it's true whenever a and b are coprime. —David Eppstein (talk) 20:26, 28 July 2016 (UTC)
 * "Trivial" is a word math instructors use as a short-hand for "I can easily see," but that to students means,"After class, go back to your desk and work for 5-75 minutes trying, sometimes unsuccessfully, to see it."2600:6C67:1C00:5F7E:6DF1:CD52:7E5F:D359 (talk) 15:41, 9 July 2022 (UTC)

Let say a and c have a common divisor, say t. Then a't + b = c't. Hence b = c't - a't which equals t(c'-a'). So t divides b as well, contrary to the condition that a and b have no common divisor (are co-prime). The same goes for b and c. So a, b and c must be pairwise co-prime. — Preceding unsigned comment added by Generisches Maskulinum (talk • contribs) 11:13, 1 December 2023 (UTC) Generisches Maskulinum (talk) 11:18, 1 December 2023 (UTC)

Argumentation of Example might be misleading
The Example states, that $$ b = 2^{6n} -1 =  64^n -1 = (64 - 1) (\cdots) = 9 \cdot 7 \cdot (\cdots)$$is divisible by 9, which ist correct (at least I could prove it by induction), but the ellipsis might be misleading, imho. Perhaps it would be beneficial to at least remark the proof by induction? Thulsadum (talk) 11:20, 28 June 2017 (UTC)

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