In Number Theory, Logic, and Cryptography

ABC Conjecture



Let A, B, and C be three coprime integers such that 

A + B = C

Now multiply together all the distinct primes that divide any of these numbers, and call the result rad(ABC).

For example, if we start with 4 + 11 = 15, we have 2 (which divides 4), 11 (which divides 11) and 3 and 5 (which divide 15), so rad(ABC) = 2 x 11 x 3 x 5 = 330.

C is almost always smaller than rad(ABC), but not always. If you start with 2 + 243 = 245, the primes are 2 (which divides 2), 3 (which divides 243), and 5 and 7 (which divide 245). So rad(ABC) = 2 x 3 x 5 x 7 = 210. In this case, C is much bigger than rad(ABC).

Let’s count C as “much bigger” whenever it’s bigger than rad(ABC)1.1 or rad(ABC)1.001 or rad(ABC)1.000000000001 or rad(ABC)1+Є. The ABC conjecture says that no matter how small Є, there will still be only finitely many examples where C counts as much bigger than rad(ABC).


The problem is to prove or disprove the conjecture.


For further information, please see:

[1]          http://mathworld.wolfram.com/abcConjecture.html

[2]          http://www.math.unicaen.fr/~nitaj/abc.html

[3]        http://en.wikipedia.org/wiki/Abc_conjecture


You can check for contributions to this problem on the solutions page.



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Tim S Roberts

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