## Task B-2.3.

Determine all integer solutions of the system of equations

$$
\begin{aligned}
& x^{2}-y^{2}-z^{2}=18 \\
& 3 x^{2}-2 y^{2}-4 z^{2}=134
\end{aligned}
$$
## Solution.

If we multiply the first equation by -3 and add it to the second equation, we get:

$$
y^{2}-z^{2}=80
$$

We conclude that $|y|>|z|$, and if we rewrite the equation in the form

$$
(y+z)(y-z)=80
$$

we can conclude that $y$ and $z$ have the same parity, because otherwise their sum and difference would be odd, and thus the product $(y+z)(y-z)$ would also be odd, which is impossible.

Therefore, $y+z$ and $y-z$ are even divisors of 80. Since $|y|>|z|$ and all ordered pairs of the form $(\pm y, \pm z)$ are solutions to equation (1), it is sufficient to consider the following cases:

- $y+z=40, y-z=2$,
- $y+z=20, y-z=4$, and
- $y+z=10, y-z=8$.

Considering all possible sign combinations, we have

$$
(y, z) \in \{(\pm 9, \pm 1), (\pm 12, \pm 8), (\pm 21, \pm 19)\}
$$

If we substitute these solutions into the equation $x^{2}=y^{2}+z^{2}+18$ from the initial system, we get $x \in \mathbb{Z}$ only when $(y, z)=(\pm 9, \pm 1)$. It holds that $x = \pm 10$.

Then all integer solutions $(x, y, z)$ are obtained by considering all possible sign combinations from $(\pm 10, \pm 9, \pm 1)$, and there are a total of $2^{3}=8$ of them.

Or listed out:

$(10,9,1),(10,9,-1),(10,-9,1),(-10,9,1),(10,-9,-1),(-10,-9,1),(-10,9,-1),(-10,-9,-1)$.