irreducibility of binomials with unity coefficientsIrreducibilityOfBinomialsWithUnityCoefficients2005-04-29 16:17:362006-12-22 16:24:57Resultirreducible polynomial% this is the default PlanetMath preamble. as your knowledge
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\newtheorem*{thmplain}{Theorem}Let $n$ be a positive integer.\, We consider the possible factorization of the binomial $x^n\!+\!1$.
\begin{itemize}
\item If $n$ has no odd prime factors, then the binomial $x^n\!+\!1$ is \PMlinkname{irreducible}{Irreducible Polynomial}.\, Thus, $x\!+\!1$, $x^2\!+\!1$, $x^4\!+\!1$, $x^8\!+\!1$ and so on are irreducible polynomials (i.e. \PMlinkescapetext{irreducible} in the field $\mathbb{Q}$ of their coefficients).\, N.B., only $x\!+\!1$ and $x^2\!+\!1$ are \PMlinkescapetext{irreducible} in the field $\mathbb{R}$; e.g. one has\, $x^4\!+\!1 = (x^2\!-\!x\sqrt{2}\!+\!1)(x^2\!+\!x\sqrt{2}\!+\!1)$.
\item If $n$ is an odd number, then $x^n\!+\!1$ is always divisible by $x\!+\!1$:
\begin{align}
x^n+1 = (x+1)(x^{n-1}-x^{n-2}+x^{n-3}-+\cdots-x+1)
\end{align}
This \PMlinkescapetext{formula} is usable when $n$ is an odd prime number, e.g.
$$x^5+1 = (x+1)(x^4-x^3+x^2-x+1).$$
\item 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:
$$x^{12}+1 = (x^4)^3+1 = (x^4+1)[(x^4)^2-x^4+1] = (x^4+1)(x^8-x^4+1)$$
\end{itemize}
There are similar results for the binomial $x^n\!+\!y^n$, and the \PMlinkescapetext{formula} corresponding to (1) is
\begin{align}
x^n+y^n = (x+y)(x^{n-1}-x^{n-2}y+x^{n-3}y^2-+\cdots-xy^{n-2}+y^n),
\end{align}
which may be verified by performing the multiplication on the right hand \PMlinkescapetext{side}.