Question 32
Determine all possible pairs of positive integers $(m, n)$ that satisfy the following:
$$
m^{10} + 23mn + n! = 2020,
$$
where $n!$ denotes the product of all positive integers from $1$ to $n$.
$$
m^{10} + 23mn + n! = 2020,
$$
where $n!$ denotes the product of all positive integers from $1$ to $n$.