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abc-conjecture

ABC Conjecture. For every $\varepsilon>0$, there exist only finitely many triples (a, b, c) of positive coprime integers, with a + b = c, such that:

$$c\gt\operatorname{rad}(abc)^{1+\varepsilon}$$

Obviously we are trying to get $\varepsilon$ as small as possible, but what if we let $\varepsilon=1$.

As (conjecturally) there are only a finite number of solutions, what is the maximum value of $c$ for $\varepsilon=1$?

Or, if this is still too hard, find $C_{max}$ for integer $\varepsilon$'s.

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The article you link to goes on to define the quality of an $abc$-triple, to be the number $q$ such that $$ c=\operatorname{rad}(abc)^q $$ So looking for triples that appear when $\epsilon=1$ is tantamount to looking for triples whose quality is greater than $2$.

The triple with the largest-known quality is said to be: $$ a=2\\ b=3^{10} \times 109\\ c=23^5 $$ which has a quality of about $1.6299$. That is, there are no known triples which satisfy the statement with $\epsilon=1$, or indeed for any $\epsilon\geq0.63$.

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  • $\begingroup$ so the $abc$-conjecture is equivalent to $q$ being bounded from above? $\endgroup$ – JMP Oct 14 '15 at 21:53
  • $\begingroup$ different question but what is the lowest known $q$? $\endgroup$ – JMP Oct 14 '15 at 21:55
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    $\begingroup$ No, it's stronger than that. The $abc$-conjecture says that not only is $q$ bounded above, there are only a finite number of triples with $q$ larger than any number greater than $1$ (whereas just knowing that $q$ is bounded above would still permit, say, there to be infinitely many triples with $q=2$). $\endgroup$ – Micah Oct 14 '15 at 22:18
  • $\begingroup$ what's wrong with $q=1$? - most i try, e.g. 2,3,5, have $q\lt1$ $\endgroup$ – JMP Oct 14 '15 at 22:20
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    $\begingroup$ Yeah, most of the time $q<1$. You can construct infinite families with $q>1$ if you're clever, but always $q$ will tend to $1$ as the numbers get larger. (At least, assuming the conjecture is true...) $\endgroup$ – Micah Oct 14 '15 at 22:21

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