PlanetMath: irreducibility of binomials with unity coefficients

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irreducibility of binomials with unity coefficients

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

If has no odd prime factors, then the binomial $x^n\!+\!1$ is irreducible. Thus, $x\!+\!1$ , $x^2\!+\!1$ , $x^4\!+\!1$ , $x^8\!+\!1$ and so on are irreducible polynomials (i.e. irreducible in the field $\mathbb{Q}$ of their coefficients). N.B., only $x\!+\!1$ and $x^2\!+\!1$ are irreducible 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 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 formula is usable when is an odd prime number, e.g.
$\displaystyle x^5+1 = (x+1)(x^4-x^3+x^2-x+1).$
When is not a prime number but has an odd prime factor , say , then we write $x^n\!+\!1 = (x^m)^p\!+\!1$ and apply the idea of (1); for example:
$\displaystyle x^{12}+1 = (x^4)^3+1 = (x^4+1)[(x^4)^2-x^4+1] = (x^4+1)(x^8-x^4+1)$

There are similar results for the binomial $x^n\!+\!y^n$ , and the formula 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 side.

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AMS MSC:	12D05 (Field theory and polynomials :: Real and complex fields :: Polynomials: factorization)
	13F15 (Commutative rings and algebras :: Arithmetic rings and other special rings :: Factorial rings, unique factorization domains)

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