How many pairs of $(x, y)$ satisfied this equation I need help to solve in $\mathbb{Z}$ the following equation
$$yx^{2}+xy^{2}=30$$ I tried to solve it by factor $30$ to $5\times 6$ and I get those two pairs  $(2, 3) \& (3, 2) $...  is their any other pairs of  $(x, y) $? ..  
 A: Factor it as $xy(x+y)=30$. Write $30=2\times3\times5$. Therefore $y$ has to be a factor of $30$. 
So $y=\pm1,\pm2,\pm3,\pm5,\pm6,\pm10,\pm15,\pm30$. 
However, if $y\geq6$ and $x\geq1$ then $xy(x+y)\geq42$. If $y\geq6$ and $x\leq-1$ then we must also have $x\leq-7$ since otherwise one factor is negative and two are positive. Then also $xy(x+y)\geq42$.


*

*$y=1$ gives $x(x+1)=30$, this gives $x=5 \vee x=-6$.

*$y=2$ gives $2x(x+2)=30$, this gives $x(x+2)=15$, thus $x=3$ or $x=-5$. 

*$y=3$ gives $3x(x+3)=30$, this gives $x(x+3)=10$, thus $x=2$ or $x=-5$. 

*$y=5$ gives $5x(x+5)=30$, this gives $x(x+5)=6$, thus $x=1$ or $x=-6$.

*$y=-1$ gives $-x(x-1)=30$, this gives $x(x-1)=-30$, which has no solutions. 

*$y=-2$ gives $-2x(x-2)=30$, this gives $x(x-2)=-15$, which has no solutions.

*$y=-3$ gives $-3x(x-3)=30$, this gives $x(x-3)=-10$, which has no solutions.

*$y=-5$ gives $-5x(x-5)=30$, this gives $x(x-5)=-6$, thus $x=2$ or $x=3$. 

*$y=-6$ gives $-6x(x-6)=30$, this gives $x(x-6)=-5$, thus $x=1$ or $x=5$. 

*$y=-10$ gives $-10x(x-10)=30$, this gives $x(x-10)=-3$, which has no real solutions.

*$y=-15$ gives $-15x(x-15)=30$, which has no real solutions.

*$y=-30$ gives $-30x(x-30)=30$,  which has no real solutions.


Conclusion: We get the pairs $(5,1), (1,5), (-6,1), (1,-6), (2,3), (3,2), \\ (2,-5), (-5,2), (5,-6), (-6,5), (3,-5), (-5,3)$
