A triple of positive integers $(a,b,c)$ is an $abc$-triple if $a$ and $b$ are coprime and $c = a + b$. Define the quality or power of an $abc$-triple as $P(a,b,c) = \frac{\log c}{\log \text{rad}(abc)}$, where $\text{rad}(k)$ denotes the product of distinct prime divisors of $k$.

One version of the $abc$-Conjecture is that for each $\varepsilon > 0$, there are finitely many $abc$-triples such that $P(a,b,c) > 1 + \varepsilon$.

There are finitely many known triples satisfying $P > 1.4$, the so called good triples, and the largest (quality) is $P(2,3^{10} \cdot 109, 23^{5}) = 1.629911684 \dots$ (discovered by E. Reyssat).

Question: Are there any known upper bounds for $P(a,b,c)$ sharper than $\log_{p^{n}} c$, where $p$ is the minimum prime dividing $abc$ and $n$ is the number of distinct prime divisors of $abc$?

Question: Is there an absolute upper bound for $P(a,b,c)$ so that no triple has higher quality?

Best Answer: If there were such a bound, asymptotic FLT would be in hand. (Thanks Ace of Base!)

  • $\begingroup$ Is there someplace that catalogs the good tirples all the way down to 1? Is it even known that there are infinitely many triples with $P>1$? $\endgroup$ – Kevin O'Bryant Nov 3 '10 at 19:50
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    $\begingroup$ It is known that there are infinitely many triples with $P > 1$; Set $a=1$, $b=3^{2^n}-1$, $c= 3^{2^n}$ and note that, by Euler's Theorem, $2^{n+1}$ divides $b$. So we now have $rad(abc) \le \dfrac{3c}{2^{n+1}} < c$. $\endgroup$ – Woett Apr 5 '11 at 19:46
  • $\begingroup$ Does Dimitrov's recent preprint at arxiv.org/pdf/1601.03572.pdf have implications for this question and asymptotic FLT? $\endgroup$ – ThiKu Jan 19 '16 at 19:05

BTW, the smallest prime dividing $abc$ is $2$.

The wikipedia page gives an estimate of the form $c \le \exp(K \cdot \mathrm{rad}(abc)^{1/3 + \epsilon})$ where it is stated that the (implied) constants are effective. This leads to an estimate $$P(a,b,c) \le \frac{C \cdot \log(c)}{\log \log(c)}$$ for $c \ge 3$ and some effectively computable constant $C$. Your "trivial" bound is not so trivial (in some sense) because it pre-supposes a knowledge of the number of distinct prime factors of $abc$.

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