Formula for $x^n+1$

Question

Is there any formula for the factorization of $x^n+1?$ for arbitrary positive real number $x$ and a positive integer $n$? Or can it be written as a sum of products of powers of $x$?

Edit: I tried to multiply it with linear polynomials $(x\pm 1)$, check some solutions modulo small primes.

Answer 1

If $n$ is odd, you always have

$$x^n+1 = (x+1)(x^{n-1} - x^{n-2} + \cdots -x +1).$$

If $n$ is even, but has an odd factor $p$, so that $n=mp$, then you have

$$x^n+1 = (x^m)^p +1$$ and you can use the formula above.

If $n=4$, you have

$$x^4+1 = x^4+2x^2+1 - 2x^2 = (x^2+1)^2 - (\sqrt{2}x)^2$$ and you can use difference of squares.

If $n$ is a higher power of $2$, then you can factor out a $4$ and use the above formula.

Answer 2

One has $x^2+1=(x+i)(x-i)$ (decomposition holds in the ring of Gaussian integers).

More generally $x^n+1$ is a factor of $x^{2n}-1$ thus it can be factorized in terms of $2n$-th roots of $1$, i.e. in some cyclotomic ring.