Talk:Abc conjecture

Categories

Categories are capitalized. To avoid red links, one should write category:Number Theory instead of category:number theory. --AugustO (talk) 16:08, 23 October 2015 (EDT)

Thanks. The formulation, although commonly reported as stated, seems slightly off - if all the numbers are positive, then of-course the ratio will be positive also.--Andy Schlafly (talk) 16:11, 23 October 2015 (EDT)

Sorry, but you are misstating the formula: it is not square-free-part(abc) x r /c... Please revisit your sources.--AugustO (talk) 16:17, 23 October 2015 (EDT)

square-free-part(abc) times (r/c)

"square-free-part(abc) times (r/c) "

That is wrong. --AugustO (talk) 16:14, 23 October 2015 (EDT)

Working Definition

The abc-conjection doesn't deal with single cases, as it is a formula which holds for "almost all" triples of numbers with certain characteristics:

Let ε > 0 be fixed. Then for almost all triplets of co-prime integers a, b and c with a+b=c holds:

$rad(abc)^{1+\epsilon }>c$

Here rad(a b c) is the "radical" of the product $a\times b\times c$, i.e., the product of all unique primes dividing $a\times b\times c$ - or, in other words, the square-free-part.

"Almost all triplets" means that there is only a finite number of triplets which violates the condition. Some observations:

• ε can be chosen as small as you like - as long it is greater than zero.
• the finite number of exceptions obviously becomes smaller, the bigger ε is chosen.
• There are infinitely many triplets violating $rad(abc)>c$ - that's why $\epsilon =0$ isn't allowed.

That's an excellent explanation, and perhaps the simplest possible. Would you like to add it to the entry, replacing the short definition that is there?--Andy Schlafly (talk) 16:54, 23 October 2015 (EDT)

Thank you! Please feel free to incorporate it! --AugustO (talk) 17:04, 23 October 2015 (EDT)