Solve the following equation on the set of positive integers:

$$
x-y-\frac{x}{y}-\frac{x^{3}}{y^{3}}+\frac{x^{4}}{y^{4}}=2017 .
$$
Let's write the positive rational number $\frac{x}{y}$ as the quotient of positive integers $a$ and $b$ that are relatively prime to each other: $\frac{x}{y}=\frac{a}{b}$. By rearranging the equation, we get

$$
x-y-2017=\frac{x}{y}+\frac{x^{3}}{y^{3}}-\frac{x^{4}}{y^{4}}
$$

which is an integer, or

$$
\frac{x}{y}+\frac{x^{3}}{y^{3}}-\frac{x^{4}}{y^{4}}=\frac{a}{b}+\frac{a^{3}}{b^{3}}-\frac{a^{4}}{b^{4}}=\frac{a b^{3}+a^{3} b-a^{4}}{b^{4}}
$$

is also an integer. Therefore, the numerator must be divisible by $b^{4}$, and thus by $b$. The first two terms of the numerator are multiples of $b$, so $a^{4}$ must also be divisible by $b$. Since $a$ and $b$ are relatively prime, this implies that $b=1$. This proves that the fraction $\frac{a}{b}$ is an integer. Let $\frac{a}{b}=\frac{x}{y}=z$ be a positive integer. Then $x=y z$, where $x, y, z$ are all positive integers. With this notation, the original equation can be written in a more manageable form:

$$
y z-y-z-z^{3}+z^{4}=2017
$$

We will now show two ways to complete the solution.

I.) Add one to both sides and factor the left side of the equation:

$$
\begin{aligned}
y z-y-z+1-z^{3}+z^{4} & =2018 \\
(y-1)(z-1)+z^{3}(z-1) & =2018 \\
(z-1)\left(y-1+z^{3}\right) & =2018
\end{aligned}
$$

The number 1009 is a prime number, so 2018 can only be factored into positive integers in two ways: $2018=1 \cdot 2018=2 \cdot 1009$. Both factors on the left side must be positive, since $z^{3}+y-1 \geq 1+1-1=1$. Furthermore, it is certain that

$$
z^{3}+y-1 \geq z+y-1>z-1
$$

so only $z^{3}+y-1=1009$ and $z-1=2$, or $z^{3}+y-1=2018$ and $z-1=1$ are possible.

From the first, $z=3, y=1009+1-27=983, x=2949$, and from the second, $z=2, y=2018+1-8=2011, x=4022$. Checking shows that both pairs of solutions satisfy the equation.

II.) First, we show that the equation has no solution for $z \geq 7$. If $z \geq 7$, then

$$
\begin{aligned}
y z-y-z-z^{3}+z^{4} & \geq 7 y-y-z-z^{3}+z^{4}>z^{4}-z^{3}-z=z\left(z^{3}-z^{2}-1\right) \geq \\
& \geq 7\left(z^{3}-z^{2}-1\right)=7 z^{2}(z-1)-7 \geq 7 \cdot 49 \cdot 6-7=2051
\end{aligned}
$$

It is immediately clear that for $z=1$, $y z-y-z-z^{3}+z^{4}=-1$, so it is sufficient to examine the cases $z=2,3,4,5,6$. From the equation, we express $y$ and check which $z$ gives an integer value for $y$:

$$
y=\frac{2017-z^{4}+z^{3}+z}{z-1}
$$

For $z=6$, the fraction is $\frac{943}{5}$, for $z=5$, the fraction is $\frac{1522}{4}$, and for $z=4$, the fraction is $\frac{1829}{3}$, none of which are integers. For $z=3$, $y=983$, and for $z=2$, $y=2011$. For the first, $x=2949$, and for the second, $x=4022$.