## Task 1 - 260931

Determine all ordered pairs $(a, b)$ of natural numbers for which the equation $2(a+$ $b)=a b$ holds!
The equation is equivalent to $4=ab-2a-2b+4=(a-2) \cdot (b-2)$, such that $a-2$ and $b-2$ have the same sign and are divisors and corresponding opposites of 4. Since $a$ and $b$ are natural numbers, it holds that $a-2 \geq -2$ and $b-2 \geq -2$, so no integer factorization of 4 with a factor of -4 is possible. Therefore, $(a-2) \cdot (b-2) = 4 \cdot 1 = (\pm 2) \cdot (\pm 2) = 1 \cdot 4$ and thus

$$
(a, b) \in \{(6,3),(0,0),(4,4),(3,6)\}
$$

That these are also solutions to the original equation is shown by verification.