ABC Conjecture: Simple example showing $\epsilon$ is necessary I was looking over Lang's discussion of the abc conjecture in his famous Algebra tome. He says

We have to give examples such that for all $C>0$ there exist natural numbers $a$,$b$, $c$ relatively prime such that $a+b=c$ and $|a|>C N_0(abc)$. But trivially, $2^n | (3^{2^n} -1)$. We consider the relations $a_n+b_n=c_n$ given by $3^{2^n}-1 = c_n$. It is clear that these relations provide the desired examples.

Well, it is not clear to me. Can someone more algebraic than I please fill me in on what he's talking about?
I understand that because $2^n|(3^{2^n}-1)$, $N_0(3^{2^n}-1) \ll 3^{2^n}-1$, but I don't see which values of $a_n$ and $b_n$ will allow us to conclude anything like $|a|>C N_0(abc)$.
 A: The $\epsilon$ is necessary in the following sense. The abc-conjecture in the first version of Oesterle says: For every $\epsilon >0$ there are only finitely many abc-triples with quality $P(a,b,c)>1+\epsilon$, where $P(a,b,c)=\log c/(\log rad (abc))$, and $a,b,c$ coprime integers with $a+b=c$.
This is wrong for $\epsilon=0$, because of the above exmaples. To simplify it,
let 
$$
(a,b,c)=(1,9^n-1,9^n)
$$
for $n\ge 1$. Then $rad(abc)=3rad(b)$, and because of $8\mid (8+1)^n-1$ we have $8\mid b$ and $4\mid b/rad(b)$, so that $rad(b)\le b/4$, and $rad(abc)=3rad(b)\le 3b/4<c$, and hence
$$
P(1,9^n-1,9^n)>1+\frac{\log(4/3)}{2n\log 3}>1+\frac{1}{8n}>1,
$$
for infinitely many abc-triples.
A: Aside: Thanks for asking this question, I was looking for Lang's examples and my copy of Algebra is at home. 
Let me give what I feel is the simplest set of examples. These are essentially due to Erdos; see Problem 60 of this book.
Let $(a_k, b_k, c_k) = \bigl(2^k(2^k -2), 1, (2^k -1)^2\bigr)$ for $k \geq 2$. Then we have $a_k + b_k  = c_k$, $\gcd(a_k,b_k,c_k) = 1$, and 
$$N_0(a_k b_k c_k) \mid (2^k - 1)(2^k - 2)< (2^k - 1)^2 = c_k.$$
In particular, $N_0(a_k b_k c_k) < c_k$ for all $k \geq 2$. This suffices to show that the $\epsilon$ in the ABC conjecture is necessary. Lang's examples are better; they tell us that $\limsup \frac{c}{N_0(abc)} = \infty$.
