$(x+y)(x^2-5xy+y^2)=-20$ $ (x+y)(x^2-5xy+y^2) \equiv (x+y)^3 \equiv 1 \pmod {7}$ so $x+y \equiv 1,2,4 \pmod{7}$ and $x+y|20$ So possible variants are :$ x+y = 2,4$ $x+y =2$ is no solution and $x+y=4 \to (x+y)^2-7xy=-5 \to xy=3 \to (x,y)=(1,3),(3,1)$