Consequences of the abc conjecture

What is the abc conjecture | Open and solved problems | Consequences | The algorithm

The abc conjecture is a very powerful conjecture and if we can prove it, we can answer a lot of open problems from it. Here you will find a list of those problems.

Fermat's last theorem

Fermat's last theorem is not an open problem, it was solved by Andrew Wiles in 1995. But it has a very complex proof, and if we can prove the abc conjecture, we might have a much simpler proof.

It seems very likely that there are no abc triples with quality greater than 2. If we can prove this, we can also prove that there are no solutions for FLT n greater than or equal to 6. Suppose we can find a solution to Aⁿ + Bⁿ = Cⁿ, with A, B, C and n positive integers and n > 5. If GCD(A, B) > 1 we divide A, B and C by GCD(A, B). Then we suppose a = Aⁿ, b = Bⁿ and c = Cⁿ. Now we have a + b = c, and GCD(a, b) = 1. We know A < B < C, so A × B × C < C³. This means q > log(Cⁿ)/log(C³) ≥ 2.

So if the strong ABC conjecture is true, we know there are finitely many abc triples with q > 2, and we only need to check those triples to make sure there are no solutions for FLT with n > 5. This leave n = 3, 4 or 5, which have been known for ages not to produce any solutions for this equation.

The set of three consequtive powerful integers is finite

Suppose we have 3 consecutive powerful integers. Powerful integers are integers which can be written as n = A²B³. If a prime p divides n, so does p². So we know that n is greater than or equal to rad(n)². If we have three consecutive powerful integers, n, n+1 and n+2, we can create the abc triple (a = 1, b = n(n + 2) = n² + 2n, c = (n + 1)² = n² + 2n + 1). We know rad(1, n(n+2), (n+1)²) ≤ 1 × √(n(n+2)) × √(n+1) < (n+1)^3/2. This means q > log((n+1)²) / log((n+1)^3/2) = 4/3

We can extend this conjecture to three powerful integers in an arithmetic sequence. Suppose we have n, n+k, n+2k with n and k coprime. We can create the abc triple k², n × (n+2k) = n² + 2kn, (n+k)² = n² + 2kn + k². The radical is less than or equal to k × √(n × (n+2k)) × √(n+k) ≤ k × (n+k)^3/2. This means q = log((n+k)²)/log(k × ((n+k)^3/2)) = log((n+k)²/(log(k) + log((n+k)^3/2). Dividing numerator and denominator by log(n+k) gives 2/(log(k)/log(n+k)+3/2). As n goes to infinity, log(k)/log(n+k) goes to 0, and we are left with q = 2/(3/2) = 4/3.

Catalan's conjecture / Mihailescu's theorem

Catalan's conjecture says that the difference of two powers can never be one, except for 3² - 2³ = 1. Preda Mihailescu already proved the theorem, but it also follows from the abc conjecture that there are only finitely many powers which differ by 1. Suppose you have A^p + 1 = B^q, and p, q > 1. The difference of two squares is 1, only for 1² - 0², so p and q cannot both be 2. We now assume a = 1, b = A^p, c = B^q. Since b or c contains a 3^rd power, rad(abc) ≤ 1 × √b × c^1/3 < c^5/6 or rad(abc) ≤ 1 × b^1/3 × √c < c^5/6. So we can create triples with quality 6/5 and if the abc conjecture is true, there are only finitely many of these.