Let's find the solutions to the equation

$$
x y + y z + z x = 80
$$

where $x, y, z$ are natural numbers, and among them, none is smaller than $x$ and none is larger than $z$.
I. solution. The requirement can also be written as:

$$
0<x \leq y \leq z
$$

Adding $x^{2}$ to both sides of (1), the left side can be factored:

$$
\left(x^{2}+x y\right)+(y z+z x)=x(x+y)+z(x+y)=(x+y)(x+z)=80+x^{2}
$$

Thus, by choosing $x$, and then decomposing $80+x^{2}$ into the product of two natural numbers, and finally taking the smaller factor (if the factors are different) as $x+y$ and the larger one as $x+z$, we get a solution, provided that the smaller factor minus $x$ is not less than $x$, in other words, if the smaller factor is at least $2x$.

For example, let $x=1$, then

$$
(y+1)(z+1)=81=1 \cdot 81=3 \cdot 27=9 \cdot 9
$$

and from the last two decompositions, we get one solution each:

$$
\text { I. } \quad x=1, \quad y=2, \quad z=26
$$

$$
\text { II. } \quad 1, \quad 8, \quad 8 \text {. }
$$

The largest possible value of $x$ is 5, because by replacing each factor on the left side of (1) with $x$, the expression decreases or remains unchanged, so $3 x^{2} \leq 80$.

Going through the values $x=2,3,4,5$, we only list the product decompositions of the right side where neither factor is less than $2x$. There is no solution from the attempt with $x=3$, because $80+3^{2}=89$ is a prime number, and from $x=5$ due to the required size relationship.

$$
\begin{array}{ll}
x=2: & 84=4 \cdot 21=6 \cdot 14=7 \cdot 12 \\
x=4: & 96=8 \cdot 12 .
\end{array}
$$

From these, the following additional solutions are obtained:

| III. | $x=2$, | $y=2$, | $z=19$; |
| ---: | ---: | ---: | ---: |
| IV. | 2, | 4, | 12 ; |
| V. | 2, | 5, | $10 ;$ |
| VI. | 4, | 4, | 8. |

Thus, the number of solutions that meet the requirement is 6.

Remark. The method used in the solution can also be discovered this way. Let's try with $x=1$. Then

$$
y+z+y z=80, \quad y=\frac{80-z}{z+1}=\frac{81-(z+1)}{z+1}=\frac{81}{z+1}-1
$$

so $z+1$ is a divisor of $81$. Generally,

$$
y=\frac{80-x z}{x+z}
$$

in order to be able to partially perform the division here, the terms of the numerator must be increased by $x^{2}$ or $z^{2}$.

II. solution. As we saw in the first solution, $1 \leq x \leq 5$; $y$ and $z$ are the two roots of the equation

$$
u^{2}-(y+z) u+y z=0
$$

for $u$, provided that these are integers not less than $x$. If we denote $y+z$ by $s$, then $y z=80-x s$, so we need to find those values of $s$ for which the

$$
u^{2}-s u+(80-x s)=0
$$

equation has roots that are integers not less than $x$. Here, $x$ can be chosen from the integers from 1 to 5.

For this, it is necessary that the discriminant of the equation be a perfect square:

$$
D=(s+2 x)^{2}-4\left(80+x^{2}\right)=t^{2}
$$

From this,

$$
\begin{gathered}
(s+2 x+t)(s+2 x-t)=4\left(80+x^{2}\right), \quad \text { and } \\
y, z=\frac{s \pm t}{2} .
\end{gathered}
$$

Since the two factors on the left side are either both even or both odd, and the right side is even, both factors must be even. Writing $80+x^{2}$ as $v \cdot w$, where $v \geq w$,

$$
\begin{array}{cl}
s+2 x+t=2 v, & s+2 x-t=2 w \\
s=v+w-2 x, & t=v-w \\
y=\frac{s-t}{2}=w-x, & z=\frac{s+t}{2}=v-x
\end{array}
$$

By trying all possible values of $x$, we get the 6 value triples found in the first solution.