1. In the domain of integers $x$, solve the equation

$$
3\left(x^{2}\right)_{5}+(3 x)_{5}=(3 x-2)(x+2)
$$

where $n_{5}$ denotes the nearest multiple of five to the number $n$, for example, $(-3)_{5}=-5$.
1. Based on the remainder when the number $x$ is divided by 5, we can distinguish five cases: (i) $x=5k$, (ii) $x=5k+1$, (iii) $x=5k+2$, (iv) $x=5k+3$ and (v) $x=5k+4$ ($k$ denotes any integer). Since the left side of the equation is clearly a multiple of five for every integer $x$, at least one of the factors $3x-2, x+2$ on the right side must be a multiple of five. The number $3x-2$ is divisible by five only for $x=5k+4$, and the number $x+2$ only for $x=5k+3$. Therefore, it suffices to analyze cases (iv) and (v) ($L$ denotes the left and $P$ the right side of the given equation):

(iv) For $x=5k+3$, we have $x^2=25k^2+30k+9$, $(x^2)_5=25k^2+30k+10$, $3x=15k+9$, $(3x)_5=15k+10$, $L=75k^2+105k+40$ and $P=75k^2+110k+35$, so from $L=P$ we get $k=1$, which corresponds to $x=5+3=8$.

(v) For $x=5k+4$, we have $x^2=25k^2+40k+16$, $(x^2)_5=25k^2+40k+15$, $3x=15k+12$, $(3x)_5=15k+10$, $L=75k^2+135k+55$ and $P=75k^2+140k+60$, so from $L=P$ we get $k=-1$, which corresponds to $x=-5+4=-1$.

Answer: The given equation has exactly two integer solutions, namely $x=-1$ and $x=8$.

For a complete solution, award 6 points (even if the student correctly analyzes all possibilities (i)-(v)), in other cases award 1 point for each correct solution, 2 points for justifying that $x$ is of the form $5k+3$ or $5k+4$.