How amazingly beautiful is it when one of the most important conjecture implies another "world famous" theorem? How often does this happen? Not so often! In this blog, I will primarily describe how \(abc\)-conjecture (a unsolved problem constructed by Masser and Oesterle in 1985), implies the "world famous" Fermat's Last Theorem, but I will also talk about some other important conjecture it implies. What is the \(abc\)-conjecture? Before stating the original statement, I will state what a radical of a integer is.
Definition (Radical). Let \(N \in \mathbb{Z}\). The radical of \(N\) is defined as the product of its unique primes factors, i.e., \[ \begin{aligned} \text{rad}(N) = \prod_{p \mid N} p \end{aligned} \] By the unique factorization theorem, we can write \[ \begin{aligned} N = p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_r^{\alpha_r} \end{aligned} \] where \(p_1, p_2, \dots, p_r\) are distinct primes numbers and \(\alpha_1, \alpha_2, \dots, \alpha_r\) are positive integer powers. Then \[ \begin{aligned} \text{rad}(N) = p_1 p_2 \cdots p_r \end{aligned} \]
Now, let's state our conjecture
Conjecture (\(abc\)). For each \(\varepsilon > 0\) there exists a constant \(\lambda_\varepsilon > 0\) such that whenever \(a, b, c \in \mathbb{Z} \setminus \{0\}\) with \(a + b = c\) and \(\gcd(a, b) = 1\), \[ \begin{aligned} \max(|a|, |b|, |c|) \le \lambda_\varepsilon \text{rad}(abc)^{1 + \varepsilon} \end{aligned} \]
At an overview, the \(abc\)-conjecture compares the radical \(\text{rad}(abc)\) of the product \(abc\) with \(c\). A trivial bound is \[ \begin{aligned} \text{rad}(abc) \le abc < c^3 \end{aligned} \] However, the conjecture asserts that we can also bound \(c\) by a power of \(\text{rad}(abc)\).
Theorem (Fermat's Last Theorem). Let \(x, y, z \in \mathbb{N}\) and \(n > 2\), then \(x^n + y^n = z^n\) has no non-trivial integer solutions.
Proof. We will prove that if \(abc\)-conjecture is true, then Fermat's Last Theorem holds. Suppose for the sake of contradiction, \(x^n + y^n = z^n\) with \(n > 2\) and \(x, y, z \in \mathbb{N}\). We may take \(\gcd(x, y) = 1\). Suppose the \(abc\)-conjecture is proved for some \(\varepsilon\). Then, \[ \begin{aligned} z^n &\le \lambda_\varepsilon \text{rad}(x^n y^n z^n)^{1 + \varepsilon} \\ &= \lambda_\varepsilon \text{rad}(xyz)^{1 + \varepsilon} \\ &\le \lambda_\varepsilon (xyz)^{1 + \varepsilon} \\ &\le \lambda_\varepsilon z^{3(1 + \varepsilon)} \end{aligned} \] For \(n > 3(1 + \varepsilon)\) \[ \begin{aligned} z \le \lambda_\varepsilon^{1/(n-3(1 + \varepsilon))} < 2 \text{ for large } n \end{aligned} \] so \(n\) is bounded above. if \(abc\)-conjecture holds for some \(\varepsilon < 1/3\), then for any \(n \ge 4 > 3(1 + \varepsilon)\), we have \(z \le \lambda_\varepsilon^{1/(n - 3(1 + \varepsilon))}\), so Fermat's Last Theorem is a finite calculation for remaining exponents. So, we proved a general bound on \(n\). Now, let's be more specific, suppose we take \(\lambda_1 = 1\) (i.e., \(\varepsilon = 1\)), such that \(\max(|a|, |b|, |c|) \le \text{rad}(abc)^2\). Then, we have \[ \begin{aligned} z^n &\le \text{rad}(x^n y^n z^n)^{2} \\ &= \text{rad}(xyz)^{2} \\ &\le (xyz)^{2} \\ &\le z^{6} \end{aligned} \] so \(n \le 6\). Thus we have Fermat's Last Theorem for \(n \ge 6\). Also, we have indiviual "elementary" (still complicated, but does not heavily rely on advanced algebraic or analytic techniques, like Taylor-Wiles' Proof) proofs by Euler (for \(n = 3\)), Fermat (for \(n = 4\)), and Legendre (for \(n = 5\)).
Let's discuss an analogous case: for \(x \ge 1\), \(\log x \le B_{\varepsilon}x^\varepsilon\) for all \(\varepsilon > 0\), but \(\log x \not\le B\). Now, an obvious question about the \(abc\)-conjecture would arise. Why can't we take \(\varepsilon = 0\)?
If \(a + b = c\) with \(\gcd(a, b) = 1\), could \(\max(|a|, |b|, |c|) \le \lambda \text{rad}(abc)\)? Let \(p\) be a prime and take \(a = 2^{p(p-1)} - 1\), \(b = 1\), \(c = 2^{p(p-1)}\). Then, \[ \begin{aligned} 2^{p-1} \equiv 1 \pmod p \implies 2^{p(p-1)} \equiv 1 \pmod p^2 \end{aligned} \] So, \(p^2 \mid a\). Thus, \[ \begin{aligned} \text{rad}(abc) = \text{rad}(a \cdot 2) \le \frac{2a}{p} \end{aligned} \] so the \(abc\)-conjecture with \(\varepsilon = 0\) would say \[ \begin{aligned} a \le \lambda \cdot \frac{2a}{p} \implies p \le 2 \lambda \end{aligned} \] This is false for large primes \(p\)!
\(abc\)-conjecture is believed to imply many of the most famous and important conjectures and theorem in diophantine geometry and algebraic number theory. I will discuss the proofs of the Catalan's Conjecture and (Weak) Hall conjecture using the \(abc\)-conjecture. It implies the following conjectures/theorems :
Conjecture (Catalan, 1844) aka Mihailescu's Theorem. The only consecutive perfect powers in \(\mathbb{Z}^+\) are \(8 = 2^3\) and \(9 = 3^2\).
Reduced to finite but impractical number of cases by Tijdeman (1974), proved by Mihailescu (2002). What does \(abc\)-conjecture say? Suppose \(x^m - y^m = 1\) in \(\mathbb{Z}^+\) where \(m, n \ge 2\). Of course \(m \ne n\), so \(\frac{1}{m} + \frac{1}{n} \le \frac{1}{2} + \frac{1}{3} = \frac{5}{6}\). Since, \(\gcd(x, y) = 1\), by \(abc\)-conjecture \[ \begin{aligned} y^n < x^m \le \lambda_{\varepsilon} \text{rad}(x^my^n)^{1 + \varepsilon} = \lambda_{\varepsilon} \text{rad}(xy)^{1 + \varepsilon} \le \lambda_{\varepsilon} (xy)^{1 + \varepsilon} \end{aligned} \] Since, \(y^n < x^m\), \(y < x^{\frac{m}{n}}\), so \[ \begin{aligned} x^m < \lambda_{\varepsilon}(x^{1 + \frac{m}{n}})^{1 + \varepsilon} = \lambda_{\varepsilon} x^{m(\frac{1}{m} + \frac{1}{n})(1 + \varepsilon)} \le \lambda_{\varepsilon} x^{m(\frac{5}{6})(1 + \varepsilon)} \end{aligned} \] Then, \[ \begin{aligned} x^{\frac{m(1 - 5 \varepsilon)}{6}} < \lambda_{\varepsilon} \implies x < \lambda_{\varepsilon}^{\frac{6}{m(1 - 5 \varepsilon)}} \text{ for } 0 < \varepsilon < \frac{1}{5} \end{aligned} \] so, \[ \begin{aligned} y < x^{\frac{m}{n}} < \lambda_{\varepsilon}^{\frac{6}{n(1 - 5 \varepsilon)}} \end{aligned} \] We have for \(0 < \varepsilon < \frac{1}{5}\) that \[ \begin{aligned} x &< \lambda_{\varepsilon}^{\frac{6}{m(1 - 5 \varepsilon)}} \\ y &< \lambda_{\varepsilon}^{\frac{6}{n(1 - 5 \varepsilon)}} \end{aligned} \] Fix \(\varepsilon\). Large \(m\) and \(n\) for \(x < 2\) and \(y < 2\), so \(x = 1\) and \(y = 1\). So \(m\) and \(n\) are bounded above, and for each \(m\) and \(n\) we have upper bounds on \(x\) and \(y\): an effectively finite number of cases to check if the \(abc\)-conjecture is proved for a specific \(\varepsilon < \frac{1}{5}\). This application doesn't follow from abc-conjecture for \(\varepsilon = 1\).
Now, we will see how \(abc\)-conjecture implies (Weak) Hall conjecture, an conjecture on Integral Solutions.
Conjecture (M. Hall, 1969). There exists a constant \(C > 0\) such that if \(y^2 = x^3 + k\) in \(\mathbb{Z}\) with \(k \ne 0\) then \(|x| \le C|k|^2\) and \(|y| \le C|k|^3\).
Conjecture (Weak Hall Conjecture). Pick \(\varepsilon > 0\). There exists a constant \(C_{\varepsilon} > 0\) such that for each \(k \in \mathbb{Z} \setminus \{0\}\), if \(y^2 = x^3 + k\) in \(\mathbb{Z}\) then \(|x| \le C_{\varepsilon}|k|^{2(1 + \varepsilon)}\) and \(|y| \le C_{\varepsilon}|k|^{3(1 + \varepsilon)}\).
We will prove the following theorem,
Theorem. If the \(abc\)-conjecture is true then for all \(\varepsilon > 0\), there exists a \(C_{\varepsilon} > 0\) such that whenever \(y^2 = x^3 + k\) in \(\mathbb{Z}\) with \(k \ne 0\) we have \[ \begin{aligned} |x| \le C_{\varepsilon}|k|^{2(1 + \varepsilon)} \text{ and } |y| \le C_{\varepsilon}|k|^{3(1 + \varepsilon)} \end{aligned} \]
Proof. May suppose \(x, y \ne 0\). Let \(d = \gcd(x^3, y^2)\),
\[
\begin{aligned}
\frac{y^2}{d} = \frac{x^3}{d} + \frac{k}{d}
\end{aligned}
\]
Set \(a = \frac{x^3}{d}\), \(b = \frac{k}{d}\), \(c = \frac{y^2}{d}\), \(R = \text{rad}(abc)\). By \(abc\)-conjecture
\[
\begin{aligned}
\frac{|x|^3}{d} &\le \lambda_{\varepsilon} R^{1 + \varepsilon} \\
\frac{|y|^2}{d} &\le \lambda_{\varepsilon} R^{1 + \varepsilon}
\end{aligned}
\]
An upper bound: \(\displaystyle R \le \prod_{p \mid ac}p \cdot \prod_{p \mid b} p \le |x||y| \cdot \text{rad}(b) \le |x||y|\frac{|k|}{d}\).
From \(\displaystyle R = \text{rad}(abc) \le \frac{|x||y||k|}{d}\),
\[
\begin{aligned}
|x|^3, |y|^2 \le d \lambda_{\varepsilon} R^{1 + \varepsilon} \le d \lambda_{\varepsilon} \left(\frac{|x||y||k|}{d}\right)^{1 + \varepsilon} < \lambda_{\varepsilon}(|x||y|)^{1 + \varepsilon}|k|^{1 + \varepsilon}
\end{aligned}
\]
Now we take cases: \(|y|^2 \le |x|^3\) or \(|x|^3 \le |y|^2\).
Case 1: If \(|y|^2 \le |x|^3\), then \(|y| \le |x|^{\frac{3}{2}}\), so
\[
\begin{aligned}
|x|^3 < \lambda_{\varepsilon}|x|^{\frac{5}{2}(1 + \varepsilon)}|k|^{1 + \varepsilon} \implies |x|^{\frac{(1 - 5 \varepsilon)}{2}} < \lambda_{\varepsilon} |k|^{1 + \varepsilon}
\end{aligned}
\]
So, for \(0 < \varepsilon < \frac{1}{5}\),
\[
\begin{aligned}
|x| < \lambda_{\varepsilon}^{\frac{2}{(1 - 5 \varepsilon)}}|k|^{\frac{2(1 + \varepsilon)}{(1 - 5 \varepsilon)}}
\end{aligned}
\]
and
\[
\begin{aligned}
|y| < |x|^{\frac{3}{2}} < \lambda_{\varepsilon}^{\frac{3}{(1 - 5 \varepsilon)}}|k|^{\frac{3(1 + \varepsilon)}{(1 - 5 \varepsilon)}}
\end{aligned}
\]
Case 2: If instead \(|x|^3 \le |y|^2\), then \(|x| \le |y|^{\frac{2}{3}}\), so
\[
\begin{aligned}
|y|^2 < \lambda_{\varepsilon} |y|^{\frac{5}{3}(1 + \varepsilon)}|k|^{1 + \varepsilon} \implies |y|^{\frac{1 - 5 \varepsilon}{3}} < \lambda_{\varepsilon} |k|^{1 + \varepsilon}
\end{aligned}
\]
so for \(\displaystyle 0 < \varepsilon < \frac{1}{5}\)
\[
\begin{aligned}
|y| < \lambda_{\varepsilon}^{\frac{3}{1 - 5 \varepsilon}}|k|^{\frac{3(1 + \varepsilon)}{(1 - 5 \varepsilon)}}
\end{aligned}
\]
and
\[
\begin{aligned}
|x| < |y|^{\frac{2}{3}} < \lambda_{\varepsilon}^{\frac{2}{1 - 5 \varepsilon}} |k|^{\frac{2(1 + \varepsilon)}{(1 - 5 \varepsilon)}}
\end{aligned}
\]
We have the same \(x\)-bound and \(y\)-bound in both cases:
\[
\begin{aligned}
|x| < \lambda_{\varepsilon}^{\frac{2}{1 - 5 \varepsilon}} |k|^{\frac{2(1 + \varepsilon)}{(1 - 5 \varepsilon)}} \\
|y| < \lambda_{\varepsilon}^{\frac{3}{1 - 5 \varepsilon}} |k|^{\frac{3(1 + \varepsilon)}{(1 - 5 \varepsilon)}}
\end{aligned}
\]
Hence proving our required result.
Set \(\displaystyle \frac{1 + \varepsilon}{1 - 5 \varepsilon} = 1 + \varepsilon'\), so \(0 < \varepsilon' < \infty\) for \(\displaystyle 0 < \varepsilon < \frac{1}{5}\) and \(\varepsilon'\) is small if and only if \(\varepsilon\) is small. Let \(\displaystyle C_{\varepsilon'} = \max \left(\lambda_{\varepsilon}^{\frac{2}{(1 - 5 \varepsilon)}}, \lambda_{\varepsilon}^{\frac{3}{(1 - 5 \varepsilon)}}\right)\).
Theorem. Assume \(abc\)-conjecture. If \(y^2 = x^3 + k\) in \(\mathbb{Z} \setminus \{0\}\) and \(\gcd(x, y) = 1\), then \[ \begin{aligned} |x| \le C_{\varepsilon} \text{rad}(k)^{2(1 + \varepsilon)} \ , \ \ |y| \le C_{\varepsilon} \text{rad}(k)^{3(1 + \varepsilon)} \ \ \ \ \ \ \ \ (1) \end{aligned} \] If \(3y^2 = x^3 + k\) in \(\mathbb{Z} \setminus \{0\}\) and \(\gcd(x, 3y) = 1\), then \[ \begin{aligned} |x| \le B_{\varepsilon} \text{rad}(k)^{2(1 + \varepsilon)} \ , \ \ |y| \le B_{\varepsilon} \text{rad}(k)^{3(1 + \varepsilon)} \ \ \ \ \ \ \ \ (2) \end{aligned} \]
These bounds use \(\text{rad}(k)\), not \(|k|\), so they're e stronger than weak Hall conjecture (but only apply when \(\gcd(x, y) = 1\) or \(\gcd(x, 3y) = 1\)).
Theorem. Equations (1) and (2) imply the abc-conjecture, and thus are together equivalent to the \(abc\)-conjecture.
This shows Mordell's equation is a far more central equation than it at first may appear to be!
Suppose \(a + b = c\) in nonzero integers with \(\gcd(a,b) = 1\). Set \[ \begin{aligned} x &= a^2 + ab + b^2 \in \mathbb{Z}^+ \\ y &= \frac{(a-b)(a+2b)(2a+b)}{2} \in \mathbb{Z}\setminus \{0\} \end{aligned} \] Then, \[ \begin{aligned} y^2 &= x^3 - 27 \left(\frac{abc}{2}\right)^2 \\ \max(|a|, |b|, |c|) &\le 2 \sqrt{x} \\ \gcd(x, y) &= 1 \text{ or } 3 \end{aligned} \] We look here just at the case \(\gcd(x, y) = 1\). From equation (1), \[ \begin{aligned} |x| \le C_{\varepsilon} \text{rad}\left(-27\left(\frac{abc}{2}\right)^2\right)^{2(1 + \varepsilon)} \le C_{\varepsilon} (3 \cdot \text{rad}(abc))^{2(1 + \varepsilon)} \end{aligned} \] so \[ \begin{aligned} \max(|a|, |b|, |c|) \le 2\sqrt{x} \le 2 C_{\varepsilon}^{\frac{1}{2}}3^{1 + \varepsilon}\text{rad}(abc)^{1 + \varepsilon} \end{aligned} \]