TITLE: Can $\frac{a^4}{b^3}$ be arbitary close to $1$?
QUESTION [8 upvotes]: I searched for positive integers $a$ and $b$ for which the difference of $a^4$ and $b^3$ is small compared to the powers $a^4$ and $b^3$. The most spectacular example I currently have is
$$9825757^4-2104527924^3=-137318688623$$ so $(a,b)=(9825757,2104527924)$ and $\frac{a^4}{b^3}=1-1.47\cdot 10^{-17}$

Questions :

Is it true that $a^4-b^3=n$ has finite many integer solutions for every integer $n\ne 0$ ? I am not sure whether Falting's theorem can be applied here.

Can $\frac{a^4}{b^3}$ be arbitary near to $1$ from above and below ? This seems to be the case, but I have no idea how to prove it.

How can I find spectacular pairs $(a,b)$ in the above sense efficiently ? I found my example by brute force.

REPLY [3 votes]: On question 2.
Taking 
$$a=k^3-3,\\b=k^4-4k,$$
for $k\in\mathbb{N}$, $k>4$, one can  construct sequence of pairs $(a,b)$, for which
$$a^4  = b^3+n,$$
where $$n = 6k^6-44k^3+81,$$
so
$$n=6a^2-8a+3<6a^2.$$
Therefore, one can construct infinitely many pairs $(a,b)$ which provide error estimation
$$\left|\dfrac{b^3}{a^4}-1\right|<\dfrac{6}{a^2}.$$

Other sequence of such kind:
$$a=(3k)^3-1,\\b=(3k)^4-4k,$$
provides better error estimation, but with the same asymptotic (~ $a^{-2}$).