1. Find the smallest possible value of the expression

$$
3 x^{2}-12 x y+y^{4} \text {, }
$$

where $x$ and $y$ are any non-negative integers.
1. Let's denote the given expression by $V$ and transform it using the method called completing the square twice:

$$
V=3 x^{2}-12 x y+y^{4}=3(x-2 y)^{2}-12 y^{2}+y^{4}=3(x-2 y)^{2}+\left(y^{2}-6\right)^{2}-36 .
$$

Clearly, $(x-2 y)^{2} \geq 0$, so the minimum value of the expression $V$ for a fixed $y$ is obtained when we set $x=2 y$. It remains to find the minimum possible value of the square $\left(y^{2}-6\right)^{2}$ with a non-negative integer variable $y$. Since $y^{2} \in\{0,1,4,9,16,25, \ldots\}$ and the number 6 falls between 4 and 9 in this set, for any integer $y$ the inequality holds:

$$
\left(y^{2}-6\right)^{2} \geq \min \left((4-6)^{2},(9-6)^{2}\right)=\min \{4,9\}=4
$$

For any integers $x$ and $y$, we thus obtain the estimate

$$
V \geq 3 \cdot 0+4-36=-32
$$

with equality $V=-32$ occurring for $y=2$ and $x=2 y=4$.

Answer. The smallest possible value of the given expression is -32.

For a complete solution, award 6 points, of which 3 points for the necessary transformation of the expression, 1 point for determining the minimum of the square $\left(y^{2}-6\right)^{2}$ (it suffices to state that $2^{2}$ is the square closest to the number 6), 1 point for determining the sought value -32, and 1 point for providing the pair $(x, y)$ for which this value is achieved.