Four, (20 points) Given that $x$ and $y$ are real numbers, and satisfy
$$
\begin{array}{l}
x y + x + y = 17, \\
x^{2} y + x y^{2} = 66 .
\end{array}
$$

Find the value of $x^{4} + x^{3} y + x^{2} y^{2} + x y^{3} + y^{4}$.
Given the conditions $x y + x + y = 17$ and $x y (x + y) = 66$, we know that $x y$ and $x + y$ are the two real roots of the equation
$$
t^{2} - 17 t + 66 = 0
$$

Solving this equation, we get
$$
t_{1} = 6, \quad t_{2} = 11.
$$

Thus, $x y = 6$ and $x + y = 11$; or $x y = 11$ and $x + y = 6$.
When $x y = 6$ and $x + y = 11$, $x$ and $y$ are the roots of the equation
$$
u^{2} - 11 u + 6 = 0
$$

Since
$$
\Delta_{1} = (-11)^{2} - 4 \times 6 = 121 - 24 > 0,
$$
the equation (2) has real roots.
$$
\begin{array}{l}
\text{At this time, } x^{2} + y^{2} = (x + y)^{2} - 2 x y \\
= 11^{2} - 2 \times 6 = 109. \\
x^{4} + x^{3} y + x^{2} y^{2} + x y^{3} + y^{4} \\
= x^{4} + y^{4} + x^{2} y^{2} + x y (x^{2} + y^{2}) \\
= (x^{2} + y^{2})^{2} - x^{2} y^{2} + x y (x^{2} + y^{2}) \\
= 109^{2} - 6^{2} + 6 \times 109 = 12499.
\end{array}
$$

When $x y = 11$ and $x + y = 6$, $x$ and $y$ are the roots of the equation
$$
v^{2} - 6 v + 11 = 0
$$

Since
$$
\Delta_{2} = (-6)^{2} - 4 \times 11 = 36 - 44 < 0,
$$
the equation (3) has no real roots.
Therefore, the value of $x^{4} + x^{3} y + x^{2} y^{2} + x y^{3} + y^{4}$ is 12499.