## Task A-4.1.

Determine all natural numbers $m$ and $n$ such that $2^{n!} = m^3 + 37$.
## Second Solution.

As in the first solution, we conclude that $n \geqslant 3$ and therefore $2^{n!}$ must be a cube of a natural number. Let's denote this cube by $N^{3}$.

If we move $m^{3}$ to the left side of the equation, we will get the difference of cubes, so we have

$$
(N-m)\left(N^{2}+N m+m^{2}\right)=37
$$

Since 37 is a prime number and $N^{2}+N m+m^{2} \geqslant 3$ (since $N$ and $m$ are natural numbers), the only possibility is

$$
N-m=1 \quad \text { and } \quad N^{2}+N m+m^{2}=37
$$

From the first equation, we get $N=m+1$, and substituting this into the second equation, we obtain

$$
(m+1)^{2}+m(m+1)+m^{2}=37, \quad \text { i.e. } \quad m^{2}+m-12=0
$$

The solutions to the above quadratic equation are 3 and -4, but since $m$ is a natural number, we only consider the solution $m=3$. For $m=3$, we have $m^{3}+37=64=2^{6}$, so $n!=6$, i.e., $n=3$.

Therefore, the only solution is $m=3, n=3$.