## Task 1.

Determine the smallest natural number $n$ for which there exist positive real numbers $x, y, z$ such that

$$
n=\lfloor x y\rfloor \cdot\lfloor y z\rfloor=(x+y+z)^{2}+2
$$

Note. For a real number $t$, we denote by $\lfloor t\rfloor$ the greatest integer less than or equal to $t$.
## Solution.

We have

$$
\lfloor x y\rfloor\lfloor y z\rfloor \leqslant x y \cdot y z \leqslant \frac{(x y+y z)^{2}}{4}
$$

where the first inequality follows from $\lfloor t\rfloor \leqslant t$, and the second is the AM-GM inequality.

Furthermore, $y(x+z) \leqslant \frac{(x+y+z)^{2}}{4}$ again by the AM-GM inequality, so combining these inequalities we get

$$
n=\lfloor x y\rfloor\lfloor y z\rfloor \leqslant \frac{(x+y+z)^{4}}{64}=\frac{(n-2)^{2}}{64}
$$

Thus, $(n-2)^{2} \geqslant 64 n$ or $4 \geqslant n(68-n)$. For $n \leqslant 67$, the right-hand side is greater than 68, so we conclude $n \geqslant 68$.

For $n=68$, we must have $\lfloor x y\rfloor \cdot\lfloor y z\rfloor=68$. But then at least one of the factors on the right-hand side is greater than or equal to 17, since 17 is the smallest divisor of 68 greater than $\sqrt{68}$. Without loss of generality, let this be $\lfloor x y\rfloor$. Then $x y \geqslant 17$, so by the AM-GM inequality, $x+y \geqslant 2 \sqrt{17}$, and thus $(x+y+z)^{2}>68$, a contradiction.

Similarly, for $n=69$, one of the numbers $\lfloor x y\rfloor,\lfloor y z\rfloor$ is greater than or equal to 23. Suppose it is $\lfloor x y\rfloor$. Then $x+y \geqslant 2 \sqrt{23}$ and $(x+y+z)^{2}>92$, a contradiction.

For $n=70$, set $y=\sqrt{17}, x=\frac{10}{\sqrt{17}}, z=\frac{7}{\sqrt{17}}$. Then $x y=10, y z=7$ and $x+y+z=2 \sqrt{17}$, so

$$
\lfloor x y\rfloor\lfloor y z\rfloor=(x+y+z)^{2}+2=70
$$

We conclude that the answer is $n=70$.