nasmith99 29.01.2016 21:26 Just expand and solve: $(x-y+z)^2=x^2-y^2+z^2$ $x^2+y^2+z^2-2xy-2yz+2zx=x^2-y^2+z^2$ $y^2-xy-yz+zx=0$ $(y-x)(y-z)=0$ So either $y=x$ or $y=z$.
