How to prove $x^{2k}+1$ is irreducible over the rational numbers I guess this is a very stupid question but I am stuck. Obviously $x^{2k}+1$ has no root in $\mathbb Q$, but I guess this does not imply it must be irreducible. Is there some very easy way to prove this?
 A: $ x^n + 1 $ is irreducible in $ \mathbf Q[x] $ precisely when $ n $ is a power of 2. Indeed, if $ n $ is not a power of 2 then write $ n = 2^{v_2(n)} k $ with $ k $ odd, then $ x^{2^{v_2(n)}} + 1 $ divides $ x^n + 1 $. To show the converse, note that $ [\mathbf Q(\zeta_{2^k}) : \mathbf Q] = \varphi(2^k) = 2^{k-1} $, so the $ 2^k $th cyclotomic polynomial is of degree $ 2^{k-1} $, however we have $ x^{2^k} - 1 = (x^{2^{k-1}} - 1)(x^{2^{k-1}} + 1) $ which implies that the $ 2^k $th cyclotomic polynomial is actually $ x^{2^{k-1}} + 1 $, and hence this polynomial is irreducible for $ k \geq 0 $.
Another argument can be obtained by noting that the prime $ 2 $ is totally ramified in the ring of integers $ \mathbf Z[\zeta_n] $, and therefore we might guess that $ x^n + 1 $ becomes Eisenstein at $ 2 $ with an appropriate variable transformation. This guess is correct: the binomial coefficient $ C(n, k) $ is always divisible by $ 2 $ for $ 1 \leq k \leq n-1 $ when $ n $ is a power of $ 2 $, therefore $ (x+1)^n + 1 $ is Eisenstein at $ 2 $, and thus irreducible in $ \mathbf Q[x] $.
A: This is false. For instance,
$$x^{12}+1= (x^4+1) (x^8-x^4+1)$$
The factorization of $x^{2k}+1$ appears in the factorization of $x^{4k}-1$, which is given by cyclotomic polynomials. 
