How do I solve this algebra problem

Question

The question goes, solve in real number.

$x^5 - 5 x^3y - 5x^2 + 5xy^2 + 5y = 16 \tag{1}$

$ y^5 + 5xy^3 + 5y^2 + 5x^2y + 5x = -57 \tag{2}$

I tried simplifying the first equation to, $$ x^5 + 5\left[ \left(xy+1 \right) \left( y - x^2 \right) \right] = 16 $$

and second equation to, $$ y^5 + 5 \left[ \left(xy+1 \right) \left( y^2 + x \right) \right] = -57$$

I know not much efforts shown, but this is where I'm stuck. Any hints on where do I go from here?

Thanks

Answer 1

By adding these two equations I ended up in a quite symmetric term

$$x^5+y^5+5[(1+xy)(y+x)(y-x+1)]~=~-41$$

I would also recommend to use the fact that you can write any $(x+y)^n$ only in terms of $xy$ and $(x+y)$ and the two $x^n,y^n$ like

$$(x+y)^3~=~x^3+y^3+3xy(x+y)$$

Maybe someone can proceed form there.

Answer 2

COMMENT.- It seems that Batominovski's comment about $(1,-2)$ is true. One has $$x^5+5(xy+1)(y-x^2)=16\\y^5+5(xy+1)(y^2+x)=-57$$ From which $$\frac{y^2+x}{y-x^2}=\frac{y^5+57}{x^5-16}\qquad(1)$$ A necessary condition for solutions $(x,y)$ is equality $(1)$. However it is not suffisant because if not it would be an infinity of solutions. Making separately

$$\frac{y^2+x}{y-x^2}=a\qquad(2)$$ $$\frac{y^5+57}{x^5-16}=a\qquad(3)$$ we have in $(2)$ a conic, hyperbola, ellipse, circle ($a=1$) and two lines ($a=-1$) while in $(3)$ there is a quintic.The corresponding graphics are suggestive in order to search solutions. In any case the graphics of the given equations (in the attached figure green and black respectively) also give $(1,-2)$ as probably only real solution.