Irreducible polynomial iff the condition is satisfied

Question

I am asked to show that the polynomial $f(x)=x^n+1 \in \mathbb{Q}[x]$ is irreducible if and only if $n=2^k$ for any integer $k\geq 0$.

Could you give me some hints what I could do to show this??

Answer 1

$\Rightarrow$ Assume by contradiction that $n \neq 2^k$. Then it has an odd divisor. Use the formula for $a^k+b^k$...

$\Leftarrow$ Use the fact that $(X^n+1)(X^n-1)=X^{2n}-1$. Write this as product of cyclotomic polynomials and use induction by $k$.