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 $x\!+\!1$ :

  (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

(2)