I cannot find the last factor of the expression $x^8-y^8$.

Question

I'm supposed to factor $x^8-y^8$ (the exponents are 8 for both if it is too difficult to see) as completely as possible. It is easy to factor this to $(x+y)(x-y)(x^2+y^2)(x^4+y^4)$. However, the book says "(Hint: there are 5 factors. Note that we sat real coefficients, not just integers)". What is this elusive 5th factor? Thanks!

Answer 1

Hint: Try to factor $x^4+y^4$ as $(x^2-axy+y^2)(x^2+axy+y^2)$.

Answer 2

Hint $x^4 + 4y^4 = (x² + 2y² + 2xy) (x² + 2y² − 2xy)$.

Answer 3

Hint $\rm\,\ x^4 + y^4 = (x^2+y^2)^2 - (\sqrt{2}\, xy)^2 =\, \cdots\ $ (factor the difference of squares)

Remark $\ $ One can perform analogous factorizations in more exotic cases, and these prove very useful for factor factoring integers having such forms. $ $ For example, $ $ Aurifeuille, Le Lasseur and Lucas $ $ discovered so-called Aurifeuillian factorizations of cyclotomic polynomials $\rm\;\Phi_n(x)\, =\, C_n(x)^2\! - n\, x\, D_n(x)^2\;$. These play a role in factoring integers of the form $\rm\; b^n \pm 1\:$, cf. the Cunningham Project. Below are some simple examples of such factorizations.

$$\begin{eqnarray} \rm x^4 + 2^2 &=\,&\rm (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\\\ \rm \frac{x^6 + 3^2}{x^2 + 3} &=\,&\rm (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\\\ \rm \frac{x^{10} - 5^5}{x^2 - 5} &=\,&\rm (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\\\ \rm \frac{x^{12} + 6^6}{x^4 + 36} &=\,&\rm (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\\\ \end{eqnarray}$$ For more on this and related topics see Sam Wagstaff's introduction to the Cunningham Project.

Answer 4

Another approach: if you know that

$$x^4+1=(x^2+\sqrt 2x+1)(x^2-\sqrt 2x+1)$$

then you can do as follows:

$$x^4+y^4=y^4\left(\frac{x^4}{y^4}+1\right)=y^4\left(\frac{x^2}{y^2}+\sqrt 2\frac xy+1\right)\left(\frac{x^2}{y^2}-\sqrt 2\frac xy+1\right)=$$

$$=\left(x^2+\sqrt 2xy+y^2\right)\left(x^2-\sqrt 2xy+y^2\right)$$