$(x^3+y^3)^2=a^2=(x+y)(x^5+y^5)\implies xy(x^2-y^2)^2=0$ So $x=0$ or $y=0$ or $x=y$ or $x+y=0$ So $a=y=y^3=y^5$ or $a=x=x^3=x^5$ or $a=2x=2x^3=x^5$ or $a=0$ So $a=-2\text{ or } -1\text{ or } 0\text{ or } 1\text{ or } 2$