On the conjecture : $|m^2-n^3|>\frac{1}{5}\sqrt[6]{m^2+n^3}$ 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?
 A: 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
$$
|m^2-n^3| < \frac{1}{52.3}\sqrt[6]{m^2+n^3}.
$$
A: As noted in the comments, this is a version of a conjecture originally made by Marshall Hall many years ago.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
$$ |m^2-n^3| > k_\epsilon \sqrt{|n|}\cdot (\log|n|)^{-c_\epsilon}. $$ 
