ABC@home is a mathematical project searching for abc-triples. An abc-triple consists of 3 numbers satisfying certain conditions. These triples might be helpful in proving the ABC-conjecture, one of the holy grails of mathematics, because of its interesting consequences.
ABC@home project URL; http://abcathome.com/
About the ABC-conjecture
An ABC-triple consists of 3 numbers a, b and c, satisfying the following conditions:
- a+b = c
- GCD(a, b) = 1
- rad(abc) < c
The GCD is the greatest common divisor, the greatest number dividing both a and b. Rad is the radical of a number. The radical is defined as the product of distinct primes dividing the number.
For example, 1, 8, 9 is an abc-triple,because:
- 1+8 = 9
- GCD(1, 8) = 1, since 1 has only one divisor
- rad(1*23*32) = 1*2*3 = 6 < 9
Now, we will assign a quality to the triple. The quality q is defined as:
- q = lograd(abc) c
This means: rad(abc)q = c
The ABC conjecture
Since rad(abc) < c, the quality will always be greater than 1. Also there are infinately many triples. This can be proven by looking at the triples 1, 9n-1, 9n. Note that 9n-1 can be divided by 8, because if 9n-1-1 is divisible by 8, so is 9n-1 = 9n-1-1*9+8. Since 91-1 is divisible by 8, so is 92-1 and so is 93-1, and so on. And since b is divisible by 8, rad(b) <= b/4. Also, the radical of c is 3 and the radical of a = 1. So rad(abc) <= 1*b/4*3 < b < c. So, this will always be an abc-triple.
So there are infinately many abc triples with q > 1. But the abc-conjecture says that there are only finitely many triples with q > 2. Or q > 1,1. Or even q > 1,0001. So
"for any e > 1, there are only finitely many abc-triples with q(abc) > e"
Consequences of the ABC conjecture
For the first consequense we do not even have to assume the abc-conjecture is true, but we'll assume the weak version is true, which says there are no triples with quality greater than 2. If this is true, one cannot find any solutions for the following equality: pn + qn = rn and n => 6. You may know this as Fermat's Last Theorem (FLT), only for n => 6 instead of n => 3. Suppose there is a solution, we can say a = pn, b=qn, and c=rn. And rad(abc) <= pqr < r3. Since c=r6, c > rad(abc)2, so q would be greater than 2. So this means we can't find solution to FLT for n => 6. Now we already know FLT is true, but we have a very complicated proof. So if we can prove this weak version of the ABC-conjecture, we have a much simpler proof of FLT, since the case for n = 3, 4 or 5 is very simple and has been known for ages.
Another consequence is that the set of 3 consecutive powerful numbers is finite. A powerful number is a number which can be written as n = a2b3. So if any prime p divides n, so does p2. A proof will follow soon here.