## Task A-3.4.

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

$$
m n^{2}=100(n+1)
$$
## First Solution.

Since $n$ and $n+1$ are relatively prime, $n^2$ divides 100. 3 points

The possibilities are $n=1, n=2, n=5$, and $n=10$.

The solutions are $(200,1),(75,2),(24,5)$, and $(11,10)$.

## Second Solution.

The given equation can be rewritten as

$$
n(m n-100)=100
$$

from which we see that $n$ must be a divisor of 100.

The number 100 has 9 natural divisors, and for each, we determine the corresponding $m$ using the formula $m=\frac{100(n+1)}{n^2}$ (X indicates that the corresponding $m$ is not a natural number):

$$
\begin{array}{c|ccccccccc}
n & 1 & 2 & 4 & 5 & 10 & 20 & 25 & 50 & 100 \\
\hline m & 200 & 75 & \mathrm{X} & 24 & 11 & \mathrm{X} & \mathrm{X} & \mathrm{X} & \mathrm{X}
\end{array}
$$

The solutions are $(200,1),(75,2),(24,5)$, and $(11,10)$.

Note: Instead of checking all cases, once we conclude that $n$ divides 100, we can proceed as follows. Let $100=n \cdot a$. Then $n(m-a)=a$, i.e., $n$ divides $a$. This means that $n^2$ divides 100. This conclusion is worth 2 points, and the solution is then completed as in the first solution.