Let x, y, a be real numbers such that $x+y=x^3+y^3=x^5+y^5=a.$Find the sum of all absolute possible values of a?

By hypothesis we have
$(x+y)(x^5+y^5)={(x^3+y^3)}^2,$

which becomes
$x^6+x{y}^5+x^{5}y+y^6=x^6+y^6+2x^3{y}^3,$
that is  $xy(x^4+y^4-2x^2{y}^2)=0 \text{or }y{(x^2-y^2)}^2=0\text{.}$
Let us discuss a few cases. If x=0, then the equations become $y=y^3=y^5=a$  We deduce that $y\in \{-1,0,1\}$ and accordingly $a\in \{-1,0,1\}\text{,}$ all of which are possible values of a. The case y=0 is similar and yields the same values of a. So assume that $xy\ne 0$ Then by the previous discussion we must have$x^2=y^2$ , that is either $x=y$ or $x=-y$ If $x=-y$ then all equations reduce to a=0, so suppose that x=y. Then the equations become $2x=2x^3=2x^5=a$ Again, we obtain $x\in \{-1,0,1\}$ and then $a\in \{-2,0,2\}\text{,}$all of which are admissible values. Hence the answer of the problem is$a\in \{-2,-1,0,1,2\}\text{.}$