On the conjecture : $|m^2-n^3|>\frac{1}{5}\sqrt[6]{m^2+n^3}$

Question

Let $m,n$ be positive integers, such that $m^2\neq n^3$. I conjecture the inequality $$|m^2-n^3|>\dfrac{1}{5}\sqrt[6]{m^2+n^3}.$$ I've tried a lot of numbers, and they all seem to work, but how do I prove it?

Answer 1

As suggested by Joe Silverman, there are counterexamples in my paper

Rational points near curves and small nonzero $|x^3-y^2|$ via lattice reduction, Lecture Notes in Computer Science 1838 (proceedings of ANTS-4, 2000; W.Bosma, ed.), 33-63. arXiv: math.NT/0005139 (https://arxiv.org/abs/math/0005139)

The best one there, which I think still holds the record for the largest ratio $n^{1/2} / |m^2-n^3|$ known, is $$ \begin{array}{rcl} m & \!\! = \!\! & 447884928428402042307918, \cr n & \!\! = \!\! & 5853886516781223, \end{array} $$ with $$ |m^2-n^3| < \frac{1}{52.3}\sqrt[6]{m^2+n^3}. $$

Answer 2

As noted in the comments, this is a version of a conjecture originally made by Marshall Hall many years ago. The original conjecture was that there is a constant $k$ so that if $m^2\ne n^3$, then $$ |m^2-n^3| > k \sqrt{|n|}. $$ As noted on Elkies' webpage (http://people.math.harvard.edu/~elkies/hall.html), this is widely believed to be false. In any case, Elkies used a clever search algorithm and a fair amount of computer time to find examples showing that $k$ would need to be quite small.

What is believed to be true is: Strong Hall Conjecture: For every $\epsilon>0$, there is a $k_\epsilon$ so that $$ |m^2-n^3| > k_\epsilon \sqrt{|n|}^{1-\epsilon} \quad\text{for all $m,n\in\mathbb Z$ with $m^2\ne n^3$}. $$ This is an easy consequence of the $ABC$-conjecture. One might ask if it's possible to replace the $\epsilon$ power with something like $$ |m^2-n^3| > k_\epsilon \sqrt{|n|}\cdot (\log|n|)^{-c_\epsilon}. $$