# Can someone explain the ABC conjecture to me?

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.

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 @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? – Joseph Skelton Sep 15 '12 at 4:46 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. – André Nicolas Sep 15 '12 at 5:20 $C_{\epsilon }$ is missing on the RHS of the first inequality (from WolframMathWorld). – Américo Tavares Sep 16 '12 at 21:54 so far, five answers, but nobody tried to explain why the abc conjecture is considered important! – kjetil b halvorsen Oct 21 '12 at 3:25

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 That pdf is really good! – Nick Alger Sep 15 '12 at 6:13

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.

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 $a+b+c=0$.

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

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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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It's always annoying how much trouble it causes when adding two things can make them bigger. :( – Hurkyl Sep 15 '12 at 11:09
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. – Byron Schmuland Sep 17 '12 at 12:42

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

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.

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 Dear Vassilis, I am downvoting this answer. Your computation of $d$ is wrong, and it seems unlikely that you've constructed a counterexample to ABC. Regards, – Matt E Oct 21 '12 at 3:29 @MattE.Dear matt can you tell me where the mistake is? – Vassilis Parassidis Oct 21 '12 at 3:57 Dear Vassililis, To take your first example, typically $3^{5*2^p}-1$ will be square free, except for a divisibility by $2^{p+2},$ $5^2$, and $11^2$. So the ratio $c/d$ will be like (some positive constant times) $2^p$, which grows much smaller than $d^{\epsilon}$, for any $\epsilon > 0$ (it is roughly $\log d$). So either you miscomputed $d$, or you misunderstood the statement of the conjecture. Regards, – Matt E Oct 21 '12 at 4:13 Dear Matt I base my conclusions on the definitions given on Wikipedia. First definition: If three positive integers a,b,c have no common factor and a+b=c and d denotes the product of the distinct prime factors of a,b,c the conjecture states d cannot be much smaller than c. Second definition: For every e>0 there are only finitely many triplets of coprime positive integers a+b=c such as that c>d^{1+e} where d denotes the product of the distinct prime factors of a,b,c. – Vassilis Parassidis Oct 21 '12 at 4:51 Dear Vassilis, The first statement is informal, and the second statement gives it a precise meaning. The point is that it won't be true that you have infinitely many examples with $c > d^{1 + \epsilon}$ (for any fixed $\epsilon$). Although your $c$ is $>$ than your $d$, the ratio goes like $\log d$, which grows smaller than $d^{\epsilon}$ for any positive $\epsilon$. Regards, – Matt E Oct 21 '12 at 13:25