I am an undergrad and I know that the conjecture may have been proven recently. But in reading about it, I am entirely confused as to what it means and why it is important. I was hoping some of you kind people could help me.

I know there are several formulations of the conjecture.

Wolfram says:

for any infinitesimal $\epsilon > 0$, there exists a constant $C_\epsilon$ such that for any three relatively prime integers $a$, $b$, $c$ satisfying $a+b=c$ the inequality $$\max (|a|, |b|, |c|) \leq C_{\epsilon}\displaystyle\prod_{p|abc} p^{1+\epsilon}$$ holds, where $p|abc$ indicates that the product is over primes $p$ which divide the product $abc$.

Then Wikipedia says:

For a positive integer $n$, the radical of $n$, denoted $\text{rad}(n)$, is the product of the distinct prime factors of $n$. If $a$, $b$, and $c$ are coprime positive integers such that $a + b = c$, it turns out that "usually" $c < \text{rad}(abc)$. The abc conjecture deals with the exceptions. Specifically, it states that for every $\epsilon>0$ there exist only finitely many triples $(a,b,c)$ of positive coprime integers with $a + b = c$ such that $$c>\text{rad}(abc)^{1+\epsilon}$$

An equivalent formulation states that for any $\epsilon > 0$, there exists a constant $K$ such that, for all triples of coprime positive integers $(a, b, c)$ satisfying $a + b = c$, the inequality $$c<K\cdot\text{rad}(abc)^{1+\epsilon}$$


A third formulation of the conjecture involves the quality $q(a, b, c)$ of the triple $(a, b, c)$, defined by: $$q(a,b,c)=\frac{\log(c)}{\log(\text{rad}(abc)}$$

I am particularly interested in the first definition, but any help with any of it would be greatly appreciated.

  • $\begingroup$ @AndréNicolas I corrected the definition. Also, I agree about the infinitesimals, I just assume it is an $\epsilon$ like in an analysis proof. It is just a positive number that we usually consider to be small, but it can truly any positive number. Is that right? $\endgroup$ – Joseph Skelton Sep 15 '12 at 4:46
  • $\begingroup$ Yes, what they should have said is that for any $\epsilon \gt 0$, there exists a $C_{\epsilon}$ such that $\dots$. The Wikipedia article is pretty good. $\endgroup$ – André Nicolas Sep 15 '12 at 5:20
  • $\begingroup$ $C_{\epsilon }$ is missing on the RHS of the first inequality (from WolframMathWorld). $\endgroup$ – Américo Tavares Sep 16 '12 at 21:54
  • $\begingroup$ so far, five answers, but nobody tried to explain why the abc conjecture is considered important! $\endgroup$ – kjetil b halvorsen Oct 21 '12 at 3:25
  • $\begingroup$ Go here $\longrightarrow$ m.youtube.com/watch?v=RkBl7WKzzRw I think you might like this YouTube channel :) $\endgroup$ – Mr Pie Feb 1 '18 at 9:52

Try Mazur's Questions about Number (1995).

One simple, surely fundamental, question has been recently asked (by Masser and Oesterle) as the distillation of some recent history of the subject, and of a good many ancient problems. This question is still unanswered, and goes under the name of the ABC-Conjecture. It has to do with the seemingly trite equation A + B + C = 0, but deals with this equation in a specially artful way.

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  • $\begingroup$ That pdf is really good! $\endgroup$ – Nick Alger Sep 15 '12 at 6:13

In Serge Lang's Algebra, he says: "One of the most fruitful analogies in mathematics is that between the integers and the ring of polynomials over a field". He then proves the abc conjecture for polynomials, and for good measure he proves Fermat's Last Theorem for polynomials. In other words, Lang is saying that if something is true for the ring of polynomials, one ought to check if it is true for that rather important ring called the integers. But it turns out that the ring of integers can be rather more troublesome, which may be surprising. So I'd say the abc conjecture is important because its proof over polynomial rings tells you it ought to be true for integers, but like Fermat it is rather more elusive than it appears. if you have access to Lang, his writeup in Chapter IV.7 is really good.

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  • 5
    $\begingroup$ It's always annoying how much trouble it causes when adding two things can make them bigger. :( $\endgroup$ – user14972 Sep 15 '12 at 11:09
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    $\begingroup$ There is also a section on the $abc$ conjecture in Math Talks for Undergraduates by Serge Lang. It has two proofs of the polynomial version. $\endgroup$ – user940 Sep 17 '12 at 12:42
  • $\begingroup$ Fermat's Last Theorem is ridiculously easy over the ring of polynomials ! Yet, it's so difficult over the integers. The abc conjecture would change that though... $\endgroup$ – Saikat Jun 20 '16 at 16:08

If one wants to avoid epsilons and constants in the formulation of the conjecture one can use this one instead.


i) $\mathrm{rad}\,(n)$ is the product of the distinct primes in $n$,

ii) $A,B,C$ are three positive coprime integers,

iii) $A+B=C\ $,

iv) $\kappa >1$,

then, with finitely many exceptions we have $$C<\mathrm{rad}\,(ABC)^{\kappa }.\tag{1}$$

For example at most finitely many instances of $C>\mathrm{rad}\,(ABC)^{1.005}$ are expected.

Addapted from The ABC-conjecture, Frits Beukers, ABC-day, Leiden, 9 September 2005.

On the other hand if one needs to find an implied or an effective constant, then the following formulation is better

For every $\varepsilon >0$ there exists $C(\varepsilon )$ such that

$$\max\left( \left\vert a\right\vert ,\left\vert b\right\vert ,\left\vert c\right\vert \right) \leq C(\varepsilon )\left( \displaystyle \prod\limits_{p\mid abc}p\right) ^{1+\varepsilon }\tag{2}$$

for all coprimes integers $a,b,c$ with $a+b+c=0$.

(From Enumerating ABC triples, Willem Jan Palenstijn, Universiteit Leiden, Universiteit Antwerpen, 26 November 2010)

Added. Here and here you can read two historical notes by Oesterlé and Masser.

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Suppose a, b, and c are coprime positive integers: $0<a<b<c=a+b$ with $\gcd(a,b)=\gcd(a,c)=\gcd(b,c)=1.$ Then (under the abc conjecture) there are only finitely many such a, b, and c such that $c>\operatorname{rad}(abc)^{1.1}$, only finitely many such that $c>\operatorname{rad}(abc)^{1.01}$, only finitely many such that $c>\operatorname{rad}(abc)^{1.001}$, etc.

Another way: let $p_1,p_2,\ldots,p_k$ be the set of primes dividing $abc$ with exponents $a_1,\ldots,a_k,b_1,\ldots,c_k$ ($\min(a_i,b_i,c_i)\ge0$ and $\max(a_i,b_i,c_i)\ge1$ for all $i$). Then $$ c=p_1^{c_1}p_2^{c_2}\cdots p_k^{c_k}>(p_1p_2\cdots p_k)^{1.001} $$ only finitely often (where 1.001 can be replaced with any number greater than 1).

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  • $\begingroup$ You forgot $a+b=c$. $\endgroup$ – Alex Becker Sep 15 '12 at 7:17
  • $\begingroup$ Right, of course. $\endgroup$ – Charles Sep 15 '12 at 7:35

Does this version qualify (from Peter Scholze's paper rejecting Shinichi Mochizuki's proof)?

enter image description here

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  • $\begingroup$ Yes. It is equivalent to $c < K \cdot rad(abc)^{1+\epsilon}$ with $K=e^{O(1)}$ $\endgroup$ – Pythagorus Jun 1 '19 at 4:21

Erica Klarreich gives a very concise and readable explanation what the ABC conjecture is about in this article in the Quanta magazine.

You may start reading (but don't have to, the article tells a thrilling story) with "The Sticking point". There you'll read

The ABC conjecture says that if you pick any exponent bigger than $1$, then there are only finitely many $a+b=c$ triples in which $c$ is larger than the product of the prime factors raised to your chosen exponent.

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