solved here

parmenides51 wrote: solved here 1. $a=0$ solution $y=-x$ 2. $a\not =0, \ x+y=a, x^{2}-xy+y^{2}=1, 3xy=a^{2}-1, $ $1=x^{4}-x^{3}y+x^{2}y^{2}-xy^{3}+y^{4}=a^{4}-5xy(x^{2}+xy+y^{2})$ $=a^{4}-5xy(a^{2}-xy)=\frac{-a^{4}+5a^{2}+5}{9}$. Therefore $(a^{2}-4)(a^{2}-1)=1$. For these cases $D=a^{2}-4xy=\frac{4-a^{2}}{3}\ge 0$. It give real roots x,y and solutions are $a=\pm 2, \ a=\pm 1$