## Task A-1.1.

Determine all pairs of natural numbers $(m, n)$ that satisfy the equation

$$
m(m-n)^{2}(m+n)=m^{4}+m n^{3}-99 n
$$
## First Solution.

By simplifying the expression, we have in sequence:

$$
\begin{aligned}
m\left(m^{2}-2 m n+n^{2}\right)(m+n) & =m\left(m^{3}+n^{3}\right)-99 n, & & \\
m\left(m^{2}-2 m n+n^{2}\right)(m+n) & =m(m+n)\left(m^{2}-m n+n^{2}\right)-99 n, & & 3 \text { points } \\
99 n & =m(m+n)(m n), & & \\
99 & =m^{2}(m+n) . & & 4 \text { points }
\end{aligned}
$$

The prime factorization of the number 99 is $99=3 \cdot 3 \cdot 11$. The only divisors of 99 that are also squares are 1 and 9. Therefore, we have two cases: if $m=1$, then $m+n=99$, i.e., $n=98$; and if $m=3$, then $m+n=11$, i.e., $n=8$.

Thus, the solutions for $(m, n)$ are $(1,98)$ and $(3,8)$.