In this problem, we wish to find all integers $m$ and $n$ that satisfy the condition $m n \geq 0$ and the equation:

$$
m^{3}+n^{3}+99 m n=33^{3}
$$

Let $s=m+n$ and $p=m n$. It is possible to express $m^{3}+n^{3}$ in terms of $s$ and $p$, using the factorization

$$
\begin{aligned}
m^{3}+n^{3} & =(m+n)\left(m^{2}-m n+n^{2}\right) \\
& =s\left(m^{2}+2 m n+n^{2}-3 m n\right) \\
& =s\left(s^{2}-3 p\right) \\
& =s^{3}-3 s p
\end{aligned}
$$

(a) Substitute the result of the expression given in terms of $s$ and $p$ into the equation we want to solve. Then, write it as a product of factors equal to zero.

(b) The product of factors is zero only when one of them is zero. Show that one of the factors being zero implies 34 solutions with $m$ and $n$ non-negative integers.

(c) For the other factor, show that it being zero is equivalent to

$$
(m-n)^{2}+(m+33)^{2}+(n+33)^{2}=0
$$

In this case, the only solution will be $m=n=-33$.
Solution

(a) Developing the equation, we have

$$
\begin{aligned}
m^{3}+n^{3}+99 m n-33^{3} & =0 \\
s^{3}-3 s p+99 p-33^{3} & =0 \\
s^{3}-33^{3}-3 p(s-33) & =0 \\
(s-33)\left(s^{2}+33 s+33^{2}\right)-3 p(s-33) & =0 \\
(s-33)\left(s^{2}+33 s-3 p+33^{2}\right) & =0
\end{aligned}
$$

(b) If $s=33$ the product will be zero. Then we have the solutions

$$
(m, n)=(0,33) ;(1,32) ;(2,31) ; \ldots ;(32,1) ;(33,0)
$$

Thus, we obtain 34 solutions to the equation.
(c) Now let's analyze when the factor $s^{2}+33 s-3 p+33^{2}$ is equal to zero.

$$
\begin{aligned}
s^{2}+33 s-3 p+33^{2} & =0 \\
2 s^{2}+66 s-6 p+2 \cdot 33^{2} & =0 \\
2(m+n)^{2}+66(m+n)-6 m n+2 \cdot 33^{2} & =0 \\
2 m^{2}+4 m n+2 n^{2}+66 m+66 n-6 m n+2 \cdot 33^{2} & =0 \\
\left(m^{2}-2 m n+n^{2}\right)+\left(m^{2}+66 m+33^{2}\right)+\left(n^{2}+66 n+33^{2}\right) & =0 \\
(m-n)^{2}+(m+33)^{2}+(n+33)^{2} & =0
\end{aligned}
$$

Since the square of a real number is always non-negative, the sum of squares equal to zero implies that each of them must be zero. Thus,

$$
\begin{aligned}
m-n & =0 \\
m+33 & =0 \\
n+33 & =0
\end{aligned}
$$

The only solution that satisfies these three conditions is $m=n=-33$. It is worth noting that this solution also satisfies the condition $m n \geq 0$.