Can $\frac{a^4}{b^3}$ be arbitary close to $1$? 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.


Regarding dot number 2 and 3:
Find such integers $a$ and $b$ so $a^4=b^3$. It's rather obvious that there are infinitely many of them. Just set $a=c^3$ and $b=c^4$ for some integer $c$. Then the ratio $\frac{(a+1)^4}{b^3}$ is arbitrarily close to $1$. Analogously, $\frac{(a-1)^4}{b^3}$ is close from below. This also gives a method of generating examples.
Edit regarding dot number one:
If $n=1$ I have found a proof that no such solution exist.
$$
a^4 - b^3 = 1 \Leftrightarrow b^3 = a^4 - 1 = (a^2 + 1)(a^2 - 1) = (a^2 + 1)(a+1)(a-1)
$$
I will show that a common divisor of any two of three factors on the right side is at most $2$. Indeed, by Euclid's algorithm
$$gcd(a-1, a+1) = gcd(a-1, 2) \in \{ 1, 2\}$$
$$gcd(a-1, a^2+1) = gcd(a-1, a^2 + 1 - 2(a-1)) =$$$$= gcd(a-1, (a-1)^2+2) = gcd(a-1, 2) \in \{ 1, 2\}$$
The third proof is analogous.
Moreover, if $a$ is even, all three of these equal $1$ while for odd $a$, all three equal $2$. If $a$ is odd, then consider $b' = \frac b 2$. Else just set $b' = b$. This yields that integral $b'^3$ is a product of coprime integers, thus all three of them are a cube. However $a+1$ and $a-1$ differ only by $2$, while $\frac{a+1}2$ and $\frac{a-1}2$ differ by $1$, therefore they can't be both cubes, since the least difference between consecutive cubes is $2^3 - 1^3 = 7$.
