prime factors of $x^n-1$

We list prime factor of the binomials$x^n-1$ in $\mathbb{Q}$, i.e. in the polynomial ring $\mathbb{Q}[x]$.  The prime factors can always be chosen to be with integer coefficients and the number of the prime factors equals to $\tau (n)$ (http://planetmath.org/TauFunction); see the proof (http://planetmath.org/FactorsOfNAndXn1).

$x-1$

$x^2-1=(x+1)(x-1)$

$x^3-1=(x^2+x+1)(x-1)$

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

$x^5-1=(x^4+x^3+x^2+x+1)(x-1)$

$x^6-1=(x^2+x+1)(x^2-x+1)(x+1)(x-1)$

$x^7-1=(x^6+x^5+x^4+x^3+x^2+x+1)(x-1)$

$x^8-1=(x^4+1)(x^2+1)(x+1)(x-1)$

$x^9-1=(x^6+x^3+1)(x^2+x+1)(x-1)$

$x^{10}-1=(x^4+x^3+x^2+1)(x^4-x^3+x^2-x+1)(x+1)(x-1)$

$x^{11}-1=(x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1)(x-1)$

$x^{12}-1=(x^4-x^2+1)(x^2+x+1)(x^2-x+1)(x^2+1)(x+1)(x-1)$

$x^{13}-1=(x^{12}+x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1)(x-1)$

$x^{14}-1=(x^6+x^5+x^4+x^3+x^2+x+1)(x^6-x^5+x^4-x^3+x^2-x+1)(x+1)(x-1)$

$x^{15}-1=(x^8-x^7+x^5-x^4+x^3-x+1)(x^4+x^3+x^2+x+1)(x^2+x+1)(x-1)$

$x^{16}-1=(x^8+1)(x^4+1)(x^2+1)(x+1)(x-1)$

$x^{17}-1=(x^{16}+x^{15}+x^{14}+\mathrm{\dots }+x^2+x+1)(x-1)$

$x^{18}-1=(x^6+x^3+1)(x^6-x^3+1)(x^2+x+1)(x^2-x+1)(x+1)(x-1)$

$x^{19}-1=(x^{18}+x^{17}+x^{16}+\mathrm{\dots }+x^2+x+1)(x-1)$

$x^{20}-1=(x^8-x^6+x^4-x^2+1)(x^4+x^3+x^2+x+1)(x^4-x^3+x^2-x+1)(x^2+1)(x+1)(x-1)$

$x^{21}-1=(x^{12}-x^{11}+x^9-x^8+x^6-x^4+x^3-x+1)(x^6+x^5+x^4+x^3+x^2+x+1)(x^2+x+1)(x-1)$

$x^{22}-1=(x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1)(x^{10}-x^9+x^8-x^7+x^6-x^5+x^4-x^3+x^2-x+1)(x+1)(x-1)$

$x^{23}-1=(x^{22}+x^{21}+x^{20}+\mathrm{\dots }+x^2+x+1)(x-1)$

$x^{24}-1=(x^8-x^4+1)(x^4-x^2+1)(x^4+1)(x^2+x+1)(x^2-x+1)(x^2+1)(x+1)(x-1)$

Note 1.  All factors shown above are irreducible polynomials (in the field  $\mathbb{Q}$  of their own coefficients), but of course they (except $x\pm 1$) may be split into factors of positive degree in certain extension fields; so e.g.

$$x^4+1=(x^2+x\sqrt{2}+1)(x^2-x\sqrt{2}+1)\mathit{\hspace{1em}}\mathrm{in}\mathrm{the}\mathrm{field}\mathbb{Q}(\sqrt{2}).$$

Note 2.  The 24 examples of factorizations are true also in the fields of characteristic $\ne 0$, but then many of the factors can be simplified or factored onwards (e.g.  $x^2+1\equiv {(x+1)}^2$  if the characteristic (http://planetmath.org/Characteristic) is 2).