## Task B-1.1.

Determine all natural numbers $x, y, z$ for which

$$
4 x^{2}+45 y^{2}+9 z^{2}-12 x y-36 y z=25
$$

where $x<y<z$.
## Solution.

If we write $45 y^{2}=9 y^{2}+36 y^{2}$, then the expression on the left side of the given equation can be written as the sum of two binomial squares:

$$
4 x^{2}-12 x y+9 y^{2}+36 y^{2}-36 y z+9 z^{2}=25
$$

Then we have

$$
(2 x-3 y)^{2}+(6 y-3 z)^{2}=25
$$

The only ways to write the number 25 as the sum of squares of two integers are:

$$
\begin{gathered}
25=( \pm 3)^{2}+( \pm 4)^{2} \\
25=0^{2}+( \pm 5)^{2}
\end{gathered}
$$

We have the following possibilities:

$$
\begin{aligned}
& 2 x-3 y= \pm 3, \quad 6 y-3 z= \pm 4 \\
& 2 x-3 y= \pm 4, \quad 6 y-3 z= \pm 3 \\
& 2 x-3 y=0, \quad 6 y-3 z= \pm 5 \\
& 2 x-3 y= \pm 5, \quad 6 y-3 z= \pm 0
\end{aligned}
$$

The Diophantine equations $6 y-3 z= \pm 4$ and $6 y-3 z= \pm 5$ have no solutions because the left side is divisible by 3, while the right side is not. Let's solve the remaining cases.

Let $2 x-3 y=-4, 6 y-3 z= \pm 3$.

Notice one particular solution: $x=1, y=2$. Then the general solution is of the form

$$
x=1+3 k, \quad y=2+2 k \text { . }
$$

Furthermore, since $x0$ and thus $2 x>3 y$, or $x>\frac{3}{2} y$. Given the initial condition $x<y$, we discard this case.

Similarly, we discard the case when $2 x-3 y=5, 6 y-3 z=0$.

Therefore, the solutions to the given equation are: $(1,2,3),(1,2,5)$ and $(2,3,6)$.