B1. Find all pairs of natural numbers $a$ and $b$ for which

$$
v=a b-2 a-4 b
$$

where $v$ is the least common multiple of $a$ and $b$.
B1. Let $d$ be the greatest common divisor of the numbers $a$ and $b$, and let $m$ and $n$ be natural numbers such that $a = d m$ and $b = d n$. Then the numbers $m$ and $n$ are coprime and it holds that $v = d m n$. Considering these relationships in the given equation, we get $d m n = d^2 m n - 2 d m - 4 d n$, from which, after dividing by $d$, we obtain

$$
m n = d m n - 2 m - 4 n
$$

The left side of the equation is divisible by $m$, so the right side must also be divisible by $m$, which means that $m$ divides $4 n$. Since $m$ and $n$ are coprime, $m$ divides 4, so $m \in \{1, 2, 4\}$. Similarly, we conclude that $n$ divides $2 m$ and thus $n$ divides 2, so $n \in \{1, 2\}$. Since $m$ and $n$ are coprime, they cannot both be even, so we have the following four possibilities for pairs $(m, n)$: $(1,1), (1,2), (2,1), (4,1)$. From the above equation, we express $d = \frac{m n + 4 n + 2 m}{m n} = 1 + \frac{4}{m} + \frac{2}{n}$ and for all four possibilities, we calculate $d = 7, 6, 5, 4$. Considering the equality $a = d m$ and $b = d n$, we get four solutions for pairs $(a, b)$, which are $(7,7), (6,12)$, $(10,5)$, and $(16,4)$.

Notation $a b = v d$ and $a = m d, b = n d$, where $m, n$ are coprime. 2 points.

Considering divisibility by $m, n$ and determining possible values of $m, n$. 2 points.

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