4. The integer solutions of the equation $x^{3}+2 x^{2} y=2009$ are
$\qquad$ .
4. $(1,1004),(-1,1005),(7,17),(-7,24)$.

From the factorization of 2009, we get
$$
\begin{array}{l}
x^{2}(x+2 y)=1^{2} \times 2009=7^{2} \times 41 \\
\Rightarrow\left\{\begin{array} { l } 
{ x ^ { 2 } = 1 , } \\
{ x + 2 y = 2 0 0 9 }
\end{array} \Rightarrow \left\{\begin{array} { l } 
{ x = 1 , } \\
{ y = 1 0 0 4 , }
\end{array} \left\{\begin{array}{l}
x=-1, \\
y=1005 ;
\end{array}\right.\right.\right.
\end{array}
$$

or $\left\{\begin{array}{l}x^{2}=7^{2}, \\ x+2 y=41\end{array} \Rightarrow\left\{\begin{array}{l}x=7, \\ y=17,\end{array}\left\{\begin{array}{l}x=-7, \\ y=24 .\end{array}\right.\right.\right.$