Editing Abc conjecture (section)

==Examples of triples with small radical==
The condition that ''ε'' > 0 is necessary as there exist infinitely many triples ''a'', ''b'', ''c'' with ''c'' > rad(''abc''). For example, let

{{block indent|<math>a = 1, \quad b = 2^{6n} - 1, \quad c = 2^{6n}, \qquad n > 1.</math>}}

The integer ''b'' is divisible by 9:

{{block indent|<math> b = 2^{6n} - 1  =  64^n - 1 = (64 - 1) (\cdots) = 9 \cdot 7 \cdot (\cdots).</math>}}

Using this fact, the following calculation is made:

{{block indent|<math>\begin{align}
\operatorname{rad}(abc) &= \operatorname{rad}(a) \operatorname{rad}(b) \operatorname{rad}(c) \\
&= \operatorname{rad}(1)  \operatorname{rad} \left ( 2^{6n} -1 \right ) \operatorname{rad} \left (2^{6n} \right ) \\
&= 2 \operatorname{rad} \left ( 2^{6n} -1 \right ) \\
&= 2 \operatorname{rad} \left ( 9 \cdot \tfrac{b}{9} \right )  \\
&\leqslant 2 \cdot 3  \cdot \tfrac{b}{9} \\
&= \tfrac{2}{3} b \\
&< \tfrac{2}{3} c.
\end{align}</math>}}

By replacing the exponent 6''n'' with other exponents forcing ''b'' to have larger square factors, the ratio between the radical and ''c'' can be made arbitrarily small. Specifically, let ''p'' > 2 be a prime and consider

{{block indent|<math>a = 1, \quad b = 2^{p(p-1)n} - 1, \quad c = 2^{p(p-1)n}, \qquad n > 1.</math>}}

Now it may be plausibly claimed that ''b'' is divisible by ''p''<sup>2</sup>:

{{block indent|<math>\begin{align}
b &= 2^{p(p-1)n} - 1 \\
&= \left(2^{p(p-1)}\right)^n - 1 \\
&= \left(2^{p(p-1)} - 1\right) (\cdots) \\
&= p^2 \cdot r (\cdots).
\end{align}</math>}}

The last step uses the fact that ''p''<sup>2</sup> divides 2<sup>''p''(''p''−1)</sup> − 1. This follows from [[Fermat's little theorem]], which shows that, for ''p'' > 2, 2<sup>''p''−1</sup> = ''pk'' + 1 for some integer ''k''. Raising both sides to the power of ''p'' then shows that 2<sup>''p''(''p''−1)</sup> = ''p''<sup>2</sup>(...) + 1.

And now with a similar calculation as above, the following results:

{{block indent|<math>\operatorname{rad}(abc) < \tfrac{2}{p} c.</math>}}

A list of the [[#Highest-quality triples|highest-quality triples]] (triples with a particularly small radical relative to ''c'') is given below; the highest quality, 1.6299, was found by Eric Reyssat {{harv|Lando|Zvonkin|2004|p=137}} for
{{block indent|1=''a'' = 2,}}
{{block indent|1=''b'' = 3<sup>10</sup>·109 = {{val|6,436,341}},}}
{{block indent|1=''c'' = 23<sup>5</sup> = {{val|6,436,343}},}}
{{block indent|1=rad(''abc'') = {{val|15042}}.}}