$f(x)=x^m+1$ is irreducible in $\mathbb{Q}[x]$ if only if $m=2^n$.

Question

I have a question

Prove that : $f(x)=x^m+1$ is irreducible in $\mathbb{Q}[x]$ if only if $m=2^n$

Thanks for your helps!

Answer 1

It's not true. If you take $f(x)=x^{6}+1$, it's irreducible in $\mathbb{Q}[x]$ but $6$ is not a power of $2$. I think you want $f(x)$ irreducible iff $m=2n$.

Answer 2

This is a consequence of the identity $$ x^{(2n+1)2^m} + 1 = (x^{2n+1} + 1) (x^{(2n+1)(2^{m} - 1)} - x^{(2n+1)(2^{m} - 2)} + \dots -1) \\= (x^{2n+1} + 1) \sum_{k=1}^{2^m}(-1)^{k-1} x^{(2n+1)(2^{m} - k)} $$