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 $\mathbb{R}$; 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 $x\!+\!1$:

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

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

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

Title irreducibility of binomials with unity coefficients IrreducibilityOfBinomialsWithUnityCoefficients 2013-03-22 15:13:08 2013-03-22 15:13:08 pahio (2872) pahio (2872) 14 pahio (2872) Result msc 12D05 msc 13F15 FactoringASumOrDifferenceOfTwoCubes PrimeFaxtorsOfXn1 PrimeFactorsOfXn1 ExpressibleInClosedForm