I don't understand the new paper, but here is the ABC conjecture:
Among triples A + B = C, with no common factors between the three, for any given epsilon > 0, there only finitely many with C > R^(1+epsilon); here, R is the "radical" of A,B,C, which is the product of the union of all primes that divide A, B, or C.
The difference between R and the product ABC is that we take out any higher powers of the primes. My intuitive interpretation is something like "in almost every irreducible A+B=C, we almost have C < product(primes(A,B,C))."
Fermat's last theorem (FLT) is a major consequence. Actually, the only proof I've seen connecting the two shows that FLT can have at most finitely many solutions. Here's a useful survey paper:
My intuitive interpretation is something like "in almost every irreducible A+B=C, we almost have C < product(primes(A,B,C))."
^^Another intuitive version the conjecture is: "if A+B=C (with no common prime factors), then it is very difficult for A, B, and C to be divisible by a prime raised to a high power." For instance, if A was divisible by 2^1000, B was divisible by 3^1000, and C by 5^1000, then these prime factors together would contribute only 235 = 30 to the "radical" R, which could allow C to perhaps be much bigger than R. This can't happen "too often" (well, maybe some finite number of times).
It was proved by Tijdeman in 1976 that the equation A + 1 = C has only finitely many solutions where A and C are both perfect powers. Think about this for a minute: they could be perfect squares, cubes, 4th powers... and perhaps A = something^1000 and C = somethingelse^1001. Probably the only obvious example of this is 2^3 + 1 = 3^2.
The ABC conjecture, if true, implies that for any positive integer k, the equation A + k = C has finitely many solutions where both A and C are perfect powers.
Among triples A + B = C, with no common factors between the three, for any given epsilon > 0, there only finitely many with C > R^(1+epsilon); here, R is the "radical" of A,B,C, which is the product of the union of all primes that divide A, B, or C.
The difference between R and the product ABC is that we take out any higher powers of the primes. My intuitive interpretation is something like "in almost every irreducible A+B=C, we almost have C < product(primes(A,B,C))."
Fermat's last theorem (FLT) is a major consequence. Actually, the only proof I've seen connecting the two shows that FLT can have at most finitely many solutions. Here's a useful survey paper:
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.24.5...
(Click on the PDF icon to download the paper for free.)