## Task 4 - 291244

Determine all triples $(x, y, z)$ of natural numbers $x, y$, and $z$ that satisfy the following system of equations (1), (2):

$$
\begin{aligned}
& x+2 y^{2}-3 z=17 \\
& x^{2}-3 y+2 z=9
\end{aligned}
$$
I. If for a triplet $(x, y, z)$ of natural numbers the system (1), (2) holds, then

$$
\begin{gathered}
3 x^{2}+2 x+4 y^{2}-9 y=61 \\
16(3 x+1)^{2}+3(8 y-9)^{2}=3187 \\
0<(3 x+1)^{2} \leq 199, \quad 0<x \leq 4
\end{gathered}
$$

For $x=1 ; 2 ; 4$, no natural number is obtained for $y$. For $x=3$, it follows from (3) that $y=4$ and from (2) that $z=6$. Therefore, only the triplet $(3,4,6)$ of natural numbers can satisfy (1), (2). Verification shows that this triplet satisfies (1), (2).

Adapted from [1]