Among real numbers, find the real solutions of: $x^2(y^2+z)=zy^2$, $y^2(z^2+x)=xz^2$, $z^2(x^2+y)=yx^2$

Question

Among real numbers, find the real solutions of the system: $x^2(y^2+z)=zy^2$, $y^2(z^2+x)=xz^2$, $z^2(x^2+y)=yx^2$.

I tried solving it as follows:

$x^2y^2+x^2z=zy^2$

$x^2y^2=z(y^2-x^2)$

$z=\frac{x^2y^2}{y^2-x^2}$

$x(z^2-y^2)=y^2z^2$

$x=\frac{y^2z^2}{z^2-y^2}$

$y=\frac{x^2z^2}{x^2-z^2}$

And here is where I got stuck. Could you please help me finish off the question?

Answer 1

Substituting $z$ we obtain two equations in $x$ and $y$, which we can solve. Then we need to consider the case $x^2=y^2$ separately.

In summary, we have only the following solutions (even over the complex numbers):
$$ (x,y,z)=(0,0,z),(0,y,0),(x,0,0). $$ So two of the variables equal to zero, and the third one arbitrary.

More precisely, the two equations are, assuming $x^2-y^2$ nonzero and substituting $z=x^2y^2(y^2-x^2)$ $$ (x^2y^2 - x^2 - xy^2 + y^2)(x^2 - xy^2 - y^2)xy^2=0, $$ and $$ (x^2y^2 + x^2y + x^2 - y^2)(x^2y - x^2 + y^2)x^2y=0. $$