T a s k 2. Find the natural roots of the equation

$17(x y z t+x y+x t+z t+1)-54(y z t+y+t)=0$.
Solution.

We solve this equation with respect to $x$:

$$
17 x=54-17(z t+1):(y z t+y+t) .
$$

From here,

$$
54-17 x=\frac{17}{y+\frac{t}{z t+1}}
$$

The right-hand side of equation (1) is an integer and positive, since $y, z, t, x$ are natural numbers. Therefore, $x \leqslant 3$. When $x=1$, the left-hand side of equation (1) equals 37, when $x=2$ it equals 20, and when $x=3$ the left-hand side equals 3. Thus, we need to solve the following equations in natural numbers:

$$
\begin{aligned}
37 & =\frac{17}{y+\frac{t}{z t+1}} \\
20 & =\frac{17}{y+\frac{t}{z t+1}} ; \\
3 & =\frac{17}{y+\frac{t}{z t+1}}
\end{aligned}
$$

From equation (2) we get:

$$
37 t:(z t+1)=17-37 y
$$

Since the right-hand side of equation (5) is negative when $y \geqslant 1$, this equation has no natural solutions.

Similarly, it can be shown that equation (3) also has no natural roots.

From equation (4) it follows that $17-3 y=3:\left(z+\frac{1}{t}\right)$. Clearly, $0<3:\left(z+\frac{1}{t}\right)<3$, where $z$ and $t$ are natural numbers. Therefore, $0<17-3 y<3$. Hence, $y=5$. Then $3:\left(z+\frac{1}{i}\right)=2$, i.e., $z+\frac{1}{t}=1+0.5$. Therefore, $z=1, t=2$.

Answer. $x=3, z=1, t=2, y=5$.