irreducibility of binomials with unity coefficients
Let $n$ be a positive integer. We consider the possible factorization of the binomial^{} ${x}^{n}+1$.

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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 ${x}^{4}+1=({x}^{2}x\sqrt{2}+1)({x}^{2}+x\sqrt{2}+1)$.

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If $n$ is an odd number^{}, then ${x}^{n}+1$ is always divisible by $x+1$:
${x}^{n}+1=(x+1)({x}^{n1}{x}^{n2}+{x}^{n3}+\mathrm{\cdots}x+1)$ (1) This is usable when $n$ is an odd prime number, e.g.
$${x}^{5}+1=(x+1)({x}^{4}{x}^{3}+{x}^{2}x+1).$$ 
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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
${x}^{n}+{y}^{n}=(x+y)({x}^{n1}{x}^{n2}y+{x}^{n3}{y}^{2}+\mathrm{\cdots}x{y}^{n2}+{y}^{n}),$  (2) 
which may be verified by performing the multiplication on the right hand .
Title  irreducibility of binomials with unity coefficients 

Canonical name  IrreducibilityOfBinomialsWithUnityCoefficients 
Date of creation  20130322 15:13:08 
Last modified on  20130322 15:13:08 
Owner  pahio (2872) 
Last modified by  pahio (2872) 
Numerical id  14 
Author  pahio (2872) 
Entry type  Result 
Classification  msc 12D05 
Classification  msc 13F15 
Related topic  FactoringASumOrDifferenceOfTwoCubes 
Related topic  PrimeFaxtorsOfXn1 
Related topic  PrimeFactorsOfXn1 
Related topic  ExpressibleInClosedForm 