4. 164 Solve the integer equation $x^{2} y+x y^{2}=30$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
［Solution］ The original equation is transformed into
$$x y(x+y)=2 \times 3 \times 5$$

By symmetry, assume $|x| \leqslant|y|$, so we have $|x| \leqslant 5$. Let $x=1$, then equation (1) becomes $y(1+y)=30$. Solving this, we get $y=5$ or -6.

Let $x=2$, then equation (1) becomes $2 y(2+y)=30$, solving this, we get $y=3$ or -5.
Let $x=3$, then equation (1) becomes $3 y(3+y)=30$, solving this, we get $y=2$ or -5.
Let $x=5$, then equation (1) becomes $5 y(5+y)=30$, solving this, we get $y=1$ or -6.
Let $x=-1$, then equation (1) becomes $-y(y-1)=30$, which has no solution.
Let $x=-2$, then equation (1) becomes $-2 y(y-2)=30$, which has no solution.
Let $x=-3$, then equation (1) becomes $-3 y(y-3)=30$, which has no solution.
Let $x=-5$, then equation (1) becomes $-5 y(y-5)=30$, solving this, we get $y=2$ or $y=3$.
Therefore, the integer solutions to the original equation are $\{(1,5) ;(5,1) ;(1,-6) ;(-6,1) ;(2,3)$; $(3,2) ;(2,-5) ;(-5,2) ;(3,-5) ;(-5,3) ;(5,-6) ;(-6,5)\}$.