You are here
Homeirreducibility of binomials with unity coefficients
Primary tabs
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. 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 $n$ is an odd number, then $x^{n}\!+\!1$ is always divisible by $x\!+\!1$:
$\displaystyle x^{n}+1=(x+1)(x^{{n1}}x^{{n2}}+x^{{n3}}+\cdotsx+1)$ (1) This formula 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 formula corresponding to (1) is
$\displaystyle x^{n}+y^{n}=(x+y)(x^{{n1}}x^{{n2}}y+x^{{n3}}y^{2}+\cdotsxy% ^{{n2}}+y^{n}),$  (2) 
which may be verified by performing the multiplication on the right hand side.
Mathematics Subject Classification
12D05 no label found13F15 no label found Forums
 Planetary Bugs
 HS/Secondary
 University/Tertiary
 Graduate/Advanced
 Industry/Practice
 Research Topics
 LaTeX help
 Math Comptetitions
 Math History
 Math Humor
 PlanetMath Comments
 PlanetMath System Updates and News
 PlanetMath help
 PlanetMath.ORG
 Strategic Communications Development
 The Math Pub
 Testing messages (ignore)
 Other useful stuff
Recent Activity
new correction: Error in proof of Proposition 2 by alex2907
Jun 24
new question: A good question by Ron Castillo
Jun 23
new question: A trascendental number. by Ron Castillo
Jun 19
new question: Banach lattice valued Bochner integrals by math ias
Jun 13
new question: young tableau and young projectors by zmth
Jun 11
new question: binomial coefficients: is this a known relation? by pfb