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The ABC conjecture stated by wikipedia says the following statements are equivalent:

I. For $\epsilon>0$, there are finite coprime triple $(a,b,c)$ satisfying $a+b=c$ such that $\mathrm{rad}(abc)^{1+\epsilon}<c$

II. For $\epsilon>0$, there exists $C_\epsilon>0$,such that for all coprime triple $(a,b,c)$ satisfying $a+b=c$, $C_{\epsilon}\mathrm{rad}(abc)^{1+\epsilon}>c$ holds.

How to prove $II \implies I$?


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Does the wikipedia article not refer to articles backing up its claims? – Rasmus Oct 2 '12 at 14:35
No maybe it is considered trivial. – tan9p Oct 3 '12 at 1:13
up vote 3 down vote accepted
  1. If $C_\epsilon < 1$ then the implication is obvious.

  2. Consider $\epsilon' = \epsilon / 2$. There exists $C_{\epsilon'}$ such that for all coprime triples with $a+b = c$, the statement $$ \mathrm{rad}(abc) > \frac{c^{1/(1+\epsilon')}}{C_{\epsilon'}^{1/(1+\epsilon')}} \tag{*}$$ is true by the ABC conjecture version (II). Now, $$ \frac{1}{1+\epsilon'} = \frac{1}{1+\epsilon} + \frac{\epsilon}{(1+\epsilon)(2+\epsilon)} $$ So if for $c$ large enough (such that $\exp\left( \frac{\epsilon}{2+2\epsilon}\log c \right) > C_{\epsilon'}$ is true), (*) implies that $$ \mathrm{rad}(abc) > \frac{c^{1/(1+\epsilon')}}{C_{\epsilon'}} = c^{1/(1+\epsilon)} \cdot \left(\frac{c^{\epsilon/(2+2\epsilon)}}{C_{\epsilon'}} \right)^{1/(1+\epsilon')} \geq c^{1/(1+\epsilon)}$$ which is precisely the statement of ABC conjecture version (I). Now using that $C_{\epsilon'}$ is a finite number, there are only finitely many $c$ for which $$ \exp\left( \frac{\epsilon}{2+2\epsilon}\log c \right) \leq C_{\epsilon'} $$ is true, and these finitely many $c$s account for the finitely many possible exceptions of ABC conjecture version (I).

The reverse implication is also simple. Let $C_\epsilon$ be defined as $$ C_\epsilon = 1 + \sup \frac{c}{\mathrm{rad}(abc)^{1+\epsilon}} $$ If (I) holds, then this sup is taken over a finite number of cases where the ABC inequality is violated, and so is finite.

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A clear proof,Thank you. – tan9p Oct 3 '12 at 1:12

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