## Task A-3.3.

Determine all pairs of natural numbers $(a, b)$ for which $a^{2} b$ divides $b^{2}+3 a$.
## First Solution.

From the given condition, we have

$$
a^{2} b k=b^{2}+3 a
$$

for some $k \in \mathbb{N}$. From $a^{2} b k-3 a=b^{2}$, it follows that $a \mid b^{2}$, and from $a^{2} b k-b^{2}=3 a$, it follows that $b \mid 3 a$. Therefore, the numbers

$$
\frac{b^{2}}{a}, \frac{3 a}{b} \quad \text { and } \quad \frac{b^{2}+3 a}{a^{2} b}=\frac{b}{a^{2}}+\frac{3}{a b}
$$

are integers. Multiplying the second and third, we get that $\frac{9 a}{b^{2}}$ is also an integer, i.e., $b^{2} \mid 9 a$. Now we have

$$
b^{2} m=9 a, a n=b^{2}
$$

for some natural numbers $m$ and $n$, from which we get $m n=9$. Thus, $n \in\{1,3,9\}$, so we see that $b^{2} \in\{a, 3 a, 9 a\}$. Substituting into the initial condition, we see that $a b \mid n+3$.

Now we consider the three possible cases for $n$.

If $n=1$, then $b^{2}=a$, and from the initial condition $b^{5} \mid 4 b^{2}$, i.e., $b^{3}$ divides 4, so $a=b=1$ is one solution.

If $n=3$, then $b^{2}=3 a$, and $\left.\frac{b^{5}}{9} \right\rvert\, 2 b^{2}$, i.e., $b^{3} \mid 18$. Since $b$ is divisible by 3, there is no solution in this case.

If $n=9$, then $b^{2}=9 a$, and $\left.\frac{b^{5}}{81} \right\rvert\, \frac{4}{3} b^{2}$, i.e., $b^{3} \mid 108$. Since $b$ is divisible by 3, we get the solution $(a, b)=(1,3)$.

The only solutions are $(1,1)$ and $(1,3)$.