3. In the set of natural numbers, solve the equation

$$
x y z+y z t+x z t+x y t=x y z t+3
$$
Solution. By dividing the equation by xyzt, we get

$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}=1+\frac{3}{x y z t}
$$

Due to symmetry, without loss of generality, we assume that

$$
x \leq y \leq z \leq t
$$

from which it follows

$$
\frac{1}{x} \geq \frac{1}{y} \geq \frac{1}{z} \geq \frac{1}{t}
$$

We get

$$
\frac{4}{x} \geq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}=1+\frac{3}{x y z t}>1
$$

from which it follows $x<4$, and since $x$ is a natural number, $x=2,3$.

1. For $x=3$, we get

$$
3\left(\frac{1}{y}+\frac{1}{z}+\frac{1}{t}\right)=1+\frac{3}{3 y z t}=1+\frac{1}{y z t}
$$

Since $y \leq z \leq t$, we have

$$
\frac{3}{y} \geq \frac{1}{y}+\frac{1}{z}+\frac{1}{t} = 1 + \frac{1}{y z t} > 1
$$

from which it follows $y<3$, and since $y$ is a natural number, $y=2$.

For $y=2$, we get

$$
3\left(\frac{1}{2}+\frac{1}{z}+\frac{1}{t}\right)=1+\frac{1}{2 z t}
$$

which simplifies to

$$
\frac{3}{2} + \frac{3}{z} + \frac{3}{t} = 1 + \frac{1}{2 z t}
$$

or

$$
\frac{3}{z} + \frac{3}{t} = -\frac{1}{2} + \frac{1}{2 z t}
$$

Since $\frac{3}{z} + \frac{3}{t} > 0$, this equation has no solution.

2. For $x=2$, we get

$$
2\left(\frac{1}{y}+\frac{1}{z}+\frac{1}{t}\right)=1+\frac{3}{2 y z t}
$$

Since $y \leq z \leq t$, we have

$$
\frac{2}{y} \geq \frac{1}{y}+\frac{1}{z}+\frac{1}{t} = 1 + \frac{3}{2 y z t} > 1
$$

from which it follows $y<2$, and since $y$ is a natural number, $y=2,3,4$.

a) For $y=4$, we get

$$
3(4 z+z t+4 t)=8 z t+3,12(z+t)=5 z t+3,12\left(\frac{1}{z}+\frac{1}{t}\right)=5+\frac{3}{z t}>5, \frac{24}{z}>5
$$

from which it follows $z \leq 4$. From (1), we get $z=4$ and the equation becomes $12(4+t)=20 t+3$, or $8 t=45$, so $t$ is not a natural number.

b) For $y=3$, we get

$$
3(3 z+z t+3 t)=6 z t+3,3(z+t)=z t+1,3\left(\frac{1}{z}+\frac{1}{t}\right)=1+\frac{1}{z x}>1, \frac{6}{z}>1, z<6
$$

from which it follows $z \leq 5$. The only possibility is $z=5$. We get $10(5+t)=15 t+3$, or $5 t=47$, so $t$ is not a natural number.

b) If $y=3$, (2) is of the form $2(3 z+z t+3 t)=3 z t+3$, or $6(z+t)=z t+3$. Then $6\left(\frac{1}{z}+\frac{1}{t}\right)=1+\frac{3}{z t}>1$, from which it follows $\frac{12}{z}>1$, or $z<12$. Possibilities for $z$ are $3,5,7,9,11$.

- If $z=3$, then $6(3+t)=3 t+3$, from which $3 t=-15$, or $t=-5 \notin \mathbb{N}$.
- If $z=5$, then $6(5+t)=5 t+3, t=-27 \notin \mathbb{N}$.
- If $z=7$, then $6(7+t)=7 t+3, t=39$.
- If $z=9$, then $6(9+t)=9 t+3$, from which $3 t=51$, or $t=17$.

Thus, in this case, the solutions are the quadruples $(2,3,7,39),(2,3,9,17)$.

Case 3. Let $x=1$. Then the equation is of the form

$$
y z+y z t+z t+y t=y z t+3
$$

or

$$
y z+z t+y t=3
$$

From (1), we get $3 y z \leq 3$, or $y z \leq 1$, from which $y=1$ and $z=1$. Then $1+2 t=3$, or $t=1$. The solution is the quadruple $(1,1,1,1)$.

Finally, the solutions to the original equation are

$$
(3,3,4,11),(3,3,5,7),(2,3,7,39),(2,3,9,17),(1,1,1,1)
$$

and all their permutations.