irreducibility of binomials with unity coefficients

Let n be a positive integer.  We consider the possible factorization of the binomialMathworldPlanetmath xn+1.

  • If n has no odd prime factors, then the binomial xn+1 is irreducible ( PolynomialMathworldPlanetmath).  Thus, x+1, x2+1, x4+1, x8+1 and so on are irreducible polynomials (i.e. in the field of their coefficientsMathworldPlanetmath).  N.B., only x+1 and x2+1 are in the field ; e.g. one has  x4+1=(x2-x2+1)(x2+x2+1).

  • If n is an odd numberMathworldPlanetmathPlanetmath, then xn+1 is always divisible by x+1:

    xn+1=(x+1)(xn-1-xn-2+xn-3-+-x+1) (1)

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

  • When n is not a prime numberMathworldPlanetmath but has an odd prime factor p, say  n=mp,  then we write  xn+1=(xm)p+1  and apply the idea of (1); for example:


There are similar results for the binomial xn+yn, and the corresponding to (1) is

xn+yn=(x+y)(xn-1-xn-2y+xn-3y2-+-xyn-2+yn), (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 2013-03-22 15:13:08
Last modified on 2013-03-22 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