irreducibility of binomials with unity coefficients

Let $n$ be a positive integer.  We consider the possible factorization of the binomial $x^n+1$.

• If $n$ has no odd prime factors, then the binomial $x^n+1$ is irreducible (http://planetmath.org/Irreducible Polynomial).  Thus, $x+1$, $x^2+1$, $x^4+1$, $x^8+1$ and so on are irreducible polynomials (i.e. in the field $\mathbb{Q}$ of their coefficients).  N.B., only $x+1$ and $x^2+1$ are in the field $ℝ$; e.g. one has  $x^4+1=(x^2-x\sqrt{2}+1)(x^2+x\sqrt{2}+1)$.

• If $n$ is an odd number, then $x^n+1$ is always divisible by $x+1$:

  $x^n+1=(x+1)(x^{n-1}-x^{n-2}+x^{n-3}-+\mathrm{\cdots }-x+1)$(1)

  This is usable when $n$ is an odd prime number, e.g.

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

• When $n$ is not a prime number but has an odd prime factor $p$, say $n=mp$,  then we write  $x^n+1={(x^m)}^p+1$  and apply the idea of (1); for example:

  $$x^{12}+1={(x^4)}^3+1=(x^4+1)[{(x^4)}^2-x^4+1]=(x^4+1)(x^8-x^4+1)$$

There are similar results for the binomial $x^n+y^n$, and the corresponding to (1) is

$x^n+y^n=(x+y)(x^{n-1}-x^{n-2}y+x^{n-3}{y}^2-+\mathrm{\cdots }-x{y}^{n-2}+y^n),$(2)

which may be verified by performing the multiplication on the right hand .