SELECTION EXAMINATION

Performing some manipulation into the first equation we find: $$\begin{gathered} \left\{\begin{array}{l}(x^2+y^2{)}^2-xy(x+y{)}^2=19\\ |x-y|=1\end{array}\right\}\iff \left\{\begin{array}{l}{x}^4+2x^2{y}^2+y^4-x^{3}y-2x^2{y}^2-x{y}^3=19\\ |x-y|=1\end{array}\right\}\iff \\ \left\{\begin{array}{l}{x}^4+y^4-x^{3}y-x{y}^3=19\\ |x-y|=1\end{array}\right\}\iff \left\{\begin{array}{l}(x-y)(x^3-y^3)=19\\ |x-y|=1\end{array}\right\}\iff \left\{\begin{array}{l}(x-y{)}^2(x^2+xy+y^2)=19\\ |x-y|=1\end{array}\right\} \\ \iff \left\{\begin{array}{l}{x}^2+xy+y^2=19\\ |x-y|=1\end{array}\right\}. \end{gathered}$$ Thus we have the following cases: $$\begin{gathered} \bullet \left\{\begin{array}{l}x-y=1\\ x^2+xy+y^2=19\end{array}\right\}\iff \left\{\begin{array}{l}y=x-1\\ x^2+xy+y^2=19\end{array}\right\}\iff \left\{\begin{array}{l}y=x-1\\ x^2+x^2-x+(x-1{)}^2=19\end{array}\right\} \\ \left\{\begin{array}{l}y=x-1\\ 3x^2-3x-18=0\end{array}\right\}\iff \left\{\begin{array}{l}y=x-1\\ x=3\text{ or }x=-2\end{array}\right\}\iff (x,y)=(3,2)\text{ or }(x,y)=(-2,-3). \end{gathered}$$ $$\begin{gathered} \bullet \left\{\begin{array}{l}x-y=-1\\ x^2+xy+y^2=19\end{array}\right\}\iff \left\{\begin{array}{l}y=x+1\\ x^2+xy+y^2=19\end{array}\right\}\iff \left\{\begin{array}{l}y=x+1\\ x^2+x^2+x+(x+1{)}^2=19\end{array}\right\} \\ \left\{\begin{array}{l}y=x+1\\ 3x^2+3x-18=0\end{array}\right\}\iff \left\{\begin{array}{l}y=x+1\\ x=-3\text{ or }x=2\end{array}\right\}\iff (x,y)=(-3,-2)\text{ or }(x,y)=(2,3). \end{gathered}$$