3
$\begingroup$

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!)

$\endgroup$
3
  • $\begingroup$ 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 $P>1$? $\endgroup$ Nov 3, 2010 at 19:50
  • 2
    $\begingroup$ 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, 2011 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, 2016 at 19:05

1 Answer 1

3
$\begingroup$

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 $abc$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.