Solving this Cubic equation

Question

$(x^2+y)(x+y^2)=(x+y)^3$ Can $x^2+y^2$ attain values $2$ and $13$? How to approach this question I tried solving this equation and couldn't solve after this: $$xy+1=3(x+y) $$

Answer 1

Develop both sides and simplify to obtain: $$x^2y^2+xy=3xy(x+y)\iff xy(xy+1)=3xy(x+y)$$ Now if $x =0$ (resp. $y=0$), the equation is satisfied for any value of $y$ (resp. $x$). The sum $x^2+y^2$ can never be equal to $2$ or $13$ since they're not squares in $\mathbf Z$.

So let us suppose $x,y\neq 0$. The above equation is equivalent to $\;xy+1=3(x+y)$.

Let $p=xy$, $s=x+y$. Then $\;x^2+y^2=s^2-2p$, so that the questions amount to the (nonlinear) systems: $$\begin{cases}p=3s-1\\s^2-2p=2 &(\text{resp.}\ 13)\end{cases}$$ The first system is equivalent to $\;p=3s-1,$ $\;s^2-6s=s(s-6)=0$. Hence:

• either $s=0,\enspace p=-1$, whence $(x,y)=(1,-1)$ or $(-1,1)$,
• or $s=6,\enspace p =17$, whence $(x,y)=\pm (1,17)$ or $\pm (17,1)$ by the second condition, but then $s$ can't be equal to $.6$

The second system is equivalent to $\;p=3s-1,$ $\;s^2-6s-11=0$, which has no integer roots: these should be divisors of $11$, i.e. $\pm 1, \pm11$,; none of which is a root.

Conclusion: The solutions of this cubic equation such that $x^2+y^2=2$ are: $$\{(1,-1), (-1,1\}.$$ There is no solution such tthat $x^2+y^2=13$.

Answer 2

$$(x^2+y)(x+y^2)=(x+y)^3\\ x^3+x^2y^2+yx+y^3 =(x+y)^3\\x^3+x^2y^2+yx+y^3=x^3+y^3+3x^2y+3xy^2\\xy+x^2y^2=3x^2y+3xy^2\\xy(1+xy)=3xy(x+y)\\$$ $$3x+3y-xy-1=0 \\y=\frac{1-3x}{3-x}=\frac{3x-1}{x-3}=\\ \frac{3(x-3)+9-1}{x-3}\\y=3+\frac{8}{x-3} $$ If x,y $\in \mathbb{Z}$ so $$x-3|8\\\frac{8}{x-3} \in \mathbb{Z} \to\\x-3=\pm1 ,\pm2 ,\pm 4 ,\pm 8$$

Answer 3

$$ xy + 1 = 3(x+y) $$ $$ xy - 3 x - 3 y + 1 = 0 $$ $$ xy - 3 x - 3 y + 9 = 8 $$ $$ (x - 3)(y-3) = 8 $$ either factor is one of $$ -8,-4,-2,-1,1,2,4,8 $$ and the other is the quotient $$ -1,-2,-4,-8,8,4,2,1 $$ so we have $x$ itself in $$ -5,-1,1,2,4,5,7,11 $$ with matching $y$ in $$ 2,1,-1,-5,11,7,5,4 $$