2. Find the smallest natural number that can be written in the form $3 a^{2}-a b^{2}-2 b-4$, where $a$ and $b$ are natural numbers.
III/2. The answer is 2. If we take $a=4$ and $b=3$, we get $3 a^{2}-a b^{2}-2 b-4=2$. It is sufficient to show that the equation $3 a^{2}-a b^{2}-2 b-4=1$ has no solutions in natural numbers. We rearrange the equation to $3 a^{2}-a b^{2}=2 b+5$. Since the right side is odd, the left side must also be odd, so $a$ must be an odd number and $b$ must be an even number. The number $b^{2}$ is therefore divisible by 4, and the remainder of $a^{2}$ when divided by 4 is 1. The left side thus has a remainder of 3 when divided by 4, while the right side has a remainder of 1, since $2 b$ is divisible by 4. We have reached a contradiction, so the equation has no solutions in natural numbers.

Answer that the number 2 can be obtained. .....................................................................

Answer that the number 2 is obtained when $a=4$ and $b=3$. .................................................

Explanation that we need to show that the number 1 cannot be obtained. .................... 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_2324ea98e20bf8c655d1g-15.jpg?height=52&width=1633&top_left_y=745&top_left_x=220)

Observation that $b$ must be an even number. ............................................... 1 point

Justification that even with the above assumption, we reach a contradiction. ........... 2 points