Mathematical Reflections 2016, Issue 5 - Problem J389

Problem:
Solve in real numbers the system of equations
$$(x^2-y+1)(y^2-x+1)=2[(x^2-y)^2+(y^2-x)^2]=4.$$

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Set $u=x^2-y$ and $v=y^2-x$. The given system can be rewritten as $$\begin{array}{rcl} u^2+v^2&=&2 \\ (u+1)(v+1)&=&4. \end{array}$$
Multiplying the second equation by $2$ and adding to the first equation, we get $$(u+v)^2+2(u+v)-8=0,$$ and solving for $u+v$, we get $u+v=-4$ or $u+v=2$. If $u+v=-4$, then $uv=4-u-v-1=7$. If $u+v=2$, then $uv=4-u-v-1=1$. We obtain the two systems of equations:
$$\begin{array}{rcl} u+v&=&-4 \\ uv&=&7, \end{array} \qquad \begin{array}{rcl} u+v&=&2 \\ uv&=&1. \end{array}$$
The first system gives no real solution. The second system gives $u=v=1$. We obtain the system of equations
$$\begin{array}{rcl} x^2-y&=&1 \\ y^2-x&=&1.\end{array}$$ Subtracting side by side the two equations, we get $(x^2-y^2)+(x-y)=0$, i.e. $$(x-y)(x+y+1)=0.$$ So, $x=y$ or $x=-1-y$. Substituting these values, we obtain $$(x,y) \in \left\{\left(\dfrac{1+\sqrt{5}}{2},\dfrac{1+\sqrt{5}}{2}\right),\left(\dfrac{1-\sqrt{5}}{2},\dfrac{1-\sqrt{5}}{2}\right),(0,-1),(-1,0),\right\}.$$