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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.
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