irreducibility of binomials with unity coefficients
Let n be a positive integer. We consider the possible factorization of the binomial xn+1.
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If n has no odd prime factors, then the binomial xn+1 is irreducible (http://planetmath.org/Irreducible Polynomial
). Thus, x+1, x2+1, x4+1, x8+1 and so on are irreducible polynomials (i.e. in the field ℚ of their coefficients
). N.B., only x+1 and x2+1 are in the field ℝ; e.g. one has x4+1=(x2-x√2+1)(x2+x√2+1).
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If n is an odd number
, 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.
x5+1=(x+1)(x4-x3+x2-x+1). -
•
When n is not a prime number
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:
x12+1=(x4)3+1=(x4+1)[(x4)2-x4+1]=(x4+1)(x8-x4+1)
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 |
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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 |