AI 中文总结
本文通过将Mordell方程解的质量分解为逼近增益与幂增益，证明弱Hall猜想等价于逼近增益的渐近上界，并推广至更一般的指数情形，同时建立与abc猜想及连分数部分商的联系。
AI 中文摘要
对于方程 $y^2=x^3+k$ 的整数解，其中 $x,y\ge 1$ 且 $k\ne 0$，整数 $x^3$、$|k|$ 和 $y^2$ 构成一个三元组，其中一项是另外两项之和。当 $\gcd(x,y)=1$ 时，该三元组互素，且 $abc$ 猜想预测其质量渐近至多为 $1$。沿用 Müller、Taktikos 和 de Weger 的方法，我们将该质量通过乘积 $P=xy|k|$ 分解为逼近增益 $G_a=\log\max(x^3,y^2)/\log P$ 和幂增益 $G_p=\log P/\log\rad(xy|k|)$，其中 $G_p$ 至少为 $1$。我们证明：弱Hall猜想（即对每个 $\delta>0$，除有限多个解外有 $|k|>x^{1/2-\delta}$）等价于渐近界 $G_a\le 1$。更一般地，对于 $6/11\le\kappa<6/5$，渐近界 $G_a\le\kappa$ 蕴含对每个 $\delta>0$，除有限多个解外有 $|k|>x^{3/\kappa-5/2-\delta}$；而对于 $3/4\le\kappa<6/5$，这两个陈述等价。由于逼近增益不超过质量，具有指数 $1\le\kappa<6/5$ 的 $abc$ 不等式对原始解给出相同界。$\kappa=1$ 的情形可推广至所有解，并恢复了从 $abc$ 猜想到弱Hall猜想的经典蕴含关系。我们还关联了 $|k|$ 的小值与 $\sqrt x$ 的大部分商。
英文摘要
For an integer solution of $y^2=x^3+k$ with $x,y\ge 1$ and $k\ne 0$, the integers $x^3$, $|k|$ and $y^2$ form a triple in which one term is the sum of the other two. When $\gcd(x,y)=1$ this triple is coprime, and the $abc$ conjecture predicts that its quality is asymptotically at most $1$. Following Müller, Taktikos and de Weger, we factor this quality through the product $P=xy|k|$ into the approximation gain $G_a=\log\max(x^3,y^2)/\log P$ and the power gain $G_p=\log P/\log\rad(xy|k|)$, which is at least $1$. We show that the weak form of Hall's conjecture, which asks that $|k|>x^{1/2-δ}$ for every $δ>0$ outside finitely many solutions, is equivalent to the asymptotic bound $G_a\le 1$. More generally, for $6/11\leκ<6/5$, an asymptotic bound $G_a\leκ$ gives $|k|>x^{3/κ-5/2-δ}$ for every $δ>0$ outside finitely many solutions, and for $3/4\leκ<6/5$ the two statements are equivalent. Since the approximation gain never exceeds the quality, an $abc$ inequality with exponent $1\leκ<6/5$ gives the same bound for primitive solutions. The case $κ=1$ extends to all solutions and recovers the classical implication from the $abc$ conjecture to weak Hall. We also relate small values of $|k|$ to large partial quotients of $\sqrt x$.
