Can someone explain the ABC conjecture to me?

Question

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 $$c<K\cdot\text{rad}(abc)^{1+\epsilon}$$

holds.

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.

Answer 1

Try Mazur's Questions about Number.

Answer 2

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

If

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.

Answer 3

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.

Answer 4

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).

Answer 5

I will present my view of the A,B,C conjecture which is based on the definitions given on Wikipedia. First I have to say that no one knows with certainty all forms of this equation. From my studies of different forms of this equation I concluded that the most crucial forms for the proof are the equations where b=1. Below I give two sequences which contradict this conjecture as presented on Wikipedia.

Let's form a sequence of the equation $a+b=c$ by setting $a=3^{5*2^p}-1$, $b=$, $c=3^{5*2^p}$ for $p\ge2$. In this sequence the values $c:d$ tend to infinity, which contradicts the first definition. We can formulate another sequence by setting $a=3^{4n}-1$, $b=1$, $c=3^{4n}$ where $n$ is an odd prime number or square of a primes except the prime $5$, for all $n>1$. Then we have the sequence $3^{12}:d=8:3$, $3^{28}:d=8:3$, $3^{36}:d=8:3$ and so on indefinitely, which contradicts the second definition. All numbers of these sequences are positive integers.