## Task A-3.2.

Determine all natural numbers $a$ and $b$ such that

$$
a^{2}=4 b+3 \cdot \text{LCM}(a, b)
$$

where $\text{LCM}(m, n)$ denotes the least common multiple of the numbers $m$ and $n$.
## Solution.

Let $d$ be the greatest common divisor of $a$ and $b$, and define

$$
p=\frac{a}{d}, \quad q=\frac{b}{d}
$$

The numbers $p$ and $q$ are relatively prime natural numbers, and we have

$$
V(a, b)=V(p d, q d)=d V(p, q)=p q d
$$

Substituting this expression into the given equation, we get

$$
\begin{aligned}
p^{2} d^{2} & =4 q d+3 p q d \\
p^{2} d & =4 q+3 p q \\
4 q & =p(p d-3 q)
\end{aligned}
$$

From the last equation, it follows that $p \mid 4 q$. Since $p$ and $q$ are relatively prime, it must be that $p \mid 4$. Therefore, we have three cases: $p=1, p=2$, or $p=4$.

Consider the first case $p=1$. Substituting, we get $d=7 q$. Direct verification shows that for any natural number $q$, the numbers $a=7 q$ and $b=7 q^{2}$ indeed satisfy the condition of the problem.

If $p=2$, substituting gives $2 d=5 q$, implying that $q$ is even. However, in this case, both $p$ and $q$ are even, which contradicts the fact that $p$ and $q$ are relatively prime. Therefore, there are no such $a$ and $b$ in this case.

If $p=4$, we get $16 d=16 q$, or $d=q$. Since $p$ and $q$ are relatively prime, $q$ must not be even. Direct verification shows that for any odd natural number $q$, the numbers $a=4 q$ and $b=q^{2}$ indeed satisfy the condition of the problem.

Finally, all solutions $(a, b)$ to the original equation are $\left(7 q, 7 q^{2}\right)$, for any natural number $q$, and $\left(4 q, q^{2}\right)$, for any odd natural number $q$.