Basic Fundamentals Part 2

Note that $\begin{aligned} x^8 - y^8 & = \; (x^4 - y^4)(x^4 + y^4) \; = \; (x^2 - y^2)(x^2 + y^2)(x^4 + y^4) \; = \; (x - y)(x + y)(x^2 + y^2)(x^4 + y^4) \\ & = \; (x - y)(x + y)(x^2 + y^2)\big((x^2+y^2)^2 - 2x^2y^2\big) \; = \; (x - y)(x + y)(x^2 + y^2)(x^2 + y^2 + \sqrt{2}xy)(x^2 + y^2 - \sqrt{2}xy) \end{aligned}$ Since $x^2 + y^2 > 0$ for all $x,y \neq 0$ , the polynomial $x^2 + y^2$ cannot be factorised over the reals. Similarly $x^2 + y^2 \pm \sqrt{2}xy = \big(x \pm \tfrac{1}{\sqrt{2}}y\big)^2 + \tfrac12y^2 > 0$ for all $x,y \neq 0$ , and so the polynomials $x^2 + y^2 \pm \sqrt{2}xy$ cannot be factorised over the reals either. Thus we end up with $\boxed{5}$ factors.