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The $abc$-conjecture states that if $a,b,c$ are positive, relatively prime integers satisfying $a+b=c$, then the product of the primes dividing $abc$ (the radical of $abc$) is $\gg_\varepsilon c^{1-\varepsilon}$ for every $\varepsilon>0$.

We know several examples of $abc$-triples that refute the stronger assertion that the radical of $abc$ is $\gg c$. One example I have seen involves taking $a=1$, $b=2^n-1$, and $c=2^n$, so that the radical of $abc$ is twice the radical of $2^n-1$. By taking $n$ highly composite - say the least common multiple of the first $k$ integers - one forces $2^n-1$ to be divisible by lots of squares of primes (those not exceeding $k$, in this case), which implies that the radical of $2^n-1$ is $\ll 2^n/n$. That is, the radical of $abc$ is $\ll c/\log c$.

I'd like to cite this example in a paper I'm writing. Can someone tell me where to find it in the literature? I'd love the original citation, but even an accessible source that explicitly works out the upper bound $\ll c/\log c$ for the radical would suffice.

(Note that I don't need to be pointed to other examples of good $abc$-triples. I've put a couple of phrases above in boldface to emphasize the specific example I'm hoping to cite.)

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I think a similar construction (maybe with different primes) appears in Bombieri's book, it have a subsection about the ABC conjecture with examples and applications. –  Asaf Sep 15 '13 at 7:16
This presentation: math.jussieu.fr/~miw/articles/pdf/abc2013VI.pdf p.3 attributes the argument to F. Beukers for 3^2^k-1, though can't find the reference. –  joro Sep 15 '13 at 10:20
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3 Answers

An argument along these lines for the triple $(1,2^n-1,2^n)$ with $n=p(p-1)$ and $p$ prime is given by Granville & Tucker, It’s as easy as abc (see first paragraph on page 1227). The radical of $abc$ in this case is bounded from above by $2b/p\simeq c/\sqrt{\log c}$, so less strong than your $c/\log c$ bound, but perhaps this does qualify as an "accessible source".

The source for Frits Beukers's proof of the $c/\log c$ bound for the triple $(1,3^{2^k}-1,3^{2^k})$, mentioned by @joro, is here.

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Although this doesn't answer the question I'm asking, I do appreciate the pointer to this paper! –  Greg Martin Sep 18 '13 at 3:39
Even though I didn't get exactly what I was looking for, I appreciate the references that I didn't know about, so Carlo earns the bounty (for this answer and his other one together). –  Greg Martin Sep 24 '13 at 17:22
thank you very much!!! –  Carlo Beenakker Sep 24 '13 at 19:03
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I searched a bit harder, and think I found the proof for the specific triple $(1,2^k-1,2^k)$ that you are looking for, in a thesis from my own university: J.P. van der Horst, Finding ABC-triples using Elliptic Curves, page 10.

The abc triple $(1,2^{k_n}-1,2^{k_n})$ is considered, with $k_n$ chosen such that $2^{k_n}=1$ (mod $q^n$). The number $q$ can be any prime number $\neq 2$.

Van der Horst proves that the integer $k_n$ exists for any $n$ and that for $n\rightarrow \infty$ the radical of $abc$ is bounded from above by $e^\delta c/\log c$, with $\delta>0$ a constant that is independent of $n$.

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I'm glad for this pointer as well! But it's still not quite the example I'm looking for (I admit it's getting harder to distinguish from what I wrote in the question). This example is a generalization of the one from your first answer (Granville/Tucker), which is the $n=2$ case. But it still deals with only one prime dividing $2^k-1$ to a high power, rather than lots of squares of different primes dividing $2^k-1$. –  Greg Martin Sep 20 '13 at 5:06
hmm, I thought the key improvement here, relative to Granville/Tucker, was the reduction of the bound from $c/\sqrt{\log c}$ down to $c/\log c$. It cannot get any lower than that, can it? –  Carlo Beenakker Sep 20 '13 at 7:13
The bound $c/\log c$ is indeed equally good in van der Horst's work as in the example I'm looking for. But I'm publishing a variant of the example I'm looking for, so I'd like to cite that specific example. (I'll include these other examples as background, though, so it's helpful to be pointed to them.) –  Greg Martin Sep 20 '13 at 16:45
Regarding "it cannot get any lower than that" - Stewart-Tijdeman have a construction that gives much better bounds on the radical, better than $c/(\log c)^A$ for any $A$. It's referenced in van der Horst's thesis, I believe. –  Greg Martin Sep 20 '13 at 16:46
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The ABC example in Martin question was given by Lang:


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that's for the abc triple $(1,3^{2^k}-1,3^{2^k})$ mentioned by @joro, right? ams.org/mathscinet-getitem?mr=1005184 –  Carlo Beenakker Sep 19 '13 at 19:43
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