2016/034) Show that $x^2-xy+y^2 - 4x - 4y + 16=0$ has only one solution in real x,y. find it

 we have multiplying by 2
$2x^2-2xy+2y^2 - 8x - 8y + 32=0$
or $(x^2 - 2xy + y^2) + (x^2-8x + 16) + (y^2 - 8y + 16) = 0$
or $(x-y)^2 +(x-4)^2 + (y-4)^2=0$
hence we have each of the squares zero giving $x=y=4$ the only solution