When Is X^N+1 Irreducible?

Any number $n$ can be uniquely written as $n=2^km,$ where $k\geq 0$ and $m$ is odd. If $m>1$ , then $x^n+1$ is reducible: $x^n+1=(x^{2^k}+1)(x^{2^k(m-1)}- x^{2^k(m-2)}+...-x^{2^k}+1)$ If $m=1$ , so $n=2^k$ , we rewrite the polynomial $x^n+1$ in the variable $y=x-1$ : $f_n(x)=(y+1)^n+1$ .

Lemma. For $n=2^k$ , $(y+1)^n\equiv y^n+1 (\pmod 2)$ (that is, all of its middle coefficients are even).

Proof. We can do it by induction on $k$ . If $k=1$ , it is obvious. If for $n=2^k$ we have $(y+1)^n\equiv y^n+1 (\pmod 2)$ , then $(y+1)^{2^{k+1}} \equiv (y^n+1)^2 \equiv y^{2n} +1 (\pmod 2)$ . The lemma is proven.

By the Eisenstein Irreducibility Criterion for prime $p=2$ , the polynomial $(y+1)^{2^k}+1$ is irreducible, so $x^{2^k}+1$ is also irreducible. There are $10$ powers of $2$ up to $1000$ : from $2^0$ to $2^9.$