Resulta $p_n(x)$ es reducible sólo cuando es "obvio". Más precisamente, el polinomio de $p_n(x)$ es reducible $\mathbb{Z}$ si y sólo si al menos una de las dos condiciones siguientes.
- Existe una extraña prime $q$ tal que $q\,\vert\,n$ $n$ $q$th poder.
- Tenemos $n = 4b^4$ algunos $b \in \mathbb{N}$.
Prueba. El "si", la dirección es fácil. En el primer caso, tenemos$$p_n(x) = x^n + n = \left(x^{n\over{q}} + \sqrt[q]{n}\right)\left(\sum_{k=0}^{q-1} (-1)^k \sqrt[q]{n^k} x^{{(q-1-k)n}\over{q}}\right),$$and in the second one, Sophie Germain's Identity gives us$$p_n(x) = x^{4b^4} + 4b^4 = \left( x^{2b^4} + 2bx^{b^4} + 2b^2\right)\left(x^{2b^4} - 2bx^{b^4} + 2b^2\right).$$Now, assume that $p_n(x)$ is reducible; this is equivalent to$$[\mathbb{Q}(\zeta_{2n}\sqrt[n]{n}) : \mathbb{Q}] < n$$because $\zeta_{2n} \sqrt[n]{n}$ is a root of $p_n(x)$. Let $c \in \mathbb{N}$ be the largest number such that $n$ is the $c$th power, say $n = t^c$. Let$$g = \text{gcd}(n, c),\text{ }a = t^{c\over{g}},$$then $n = a^g$.
Necesitamos estudiar el campo $\mathbb{Q}(\sqrt[n]{n})$. Puesto que todas las raíces de $x^n - n$ han módulo de $\sqrt[n]{n}$, el módulo de la constante término del polinomio mínimo de a $\sqrt[n]{n}$ $\mathbb{Q}$ es $$\left(\sqrt[n]{n}\right)^{[\mathbb{Q}(\sqrt[n]{n}) : \mathbb{Q}]} = t^{{[\mathbb{Q}(\sqrt[n]{n}): \mathbb{Q}] \cdot c}\over{n}}.$$This field is real, and so $|x| = \pm x$ for every $x \in \mathbb{Q}$, thus$$t^{{[\mathbb{Q}(\sqrt[n]{n}): \mathbb{Q}] \cdot c}\over{n}} \in \mathbb{Q}.$$Since $t$ is not a perfect power, this implies $${{[\mathbb{Q}(\sqrt[n]{n}): \mathbb{Q}] \cdot c}\over{n}}\in\mathbb{N},$$so $$n\text{ divides }{{[\mathbb{Q}(\sqrt[n]{n}): \mathbb{Q}] \cdot c}},$$so $${n\over{g}}\text{ divides }{{[\mathbb{Q}(\sqrt[n]{n}):\mathbb{Q}]\cdot c}\over{g}}.$$Since$$\gcd\left({n\over{g}},{c\over{g}}\right)=1,$$we get $${n\over{g}}\text{ divides }[\mathbb{Q}(\sqrt[n]{n}):\mathbb{Q}],$$and hence $$[\mathbb{Q}(\sqrt[n]{n}):\mathbb{Q}]\ge {n\over{g}}.$$On the other hand, $\sqrt[n]{n}=\sqrt[n/g]{a}$, so $$[\mathbb{Q}(\sqrt[n]{n}):\mathbb{Q}]=[\mathbb{Q}(\sqrt[n/g]{a}):\mathbb{Q}]\le {n\over{g}},$$hence together, we get $$[\mathbb{Q}(\sqrt[n]{n}):\mathbb{Q}]= {n\over{g}}.$$(Notice that this implies that the polynomial $x^n-n$ is reducible if and only if $g>1$; in other words, if and only if there exists a prime $q$ such that $q\a mediados de n$ and $n$ is the $q$th power). The field$$\mathbb{Q}(\sqrt[n]{n})=\mathbb{Q}(\sqrt[n/g]{a})$$has a subfield $\mathbb{Q}(\sqrt[d]{a})$ for every $d\mid(n/g)$, and similar calculation as above shows$$[\mathbb{Q}(\sqrt[d]{a}): \mathbb{Q}]=d.$$These are in fact the only subfields of $\mathbb{Q}(\sqrt[n/g]{a})$: if $F$ is any subfield of $\mathbb{Q}(\sqrt[n/g]{a})$, then $$d:=[F:\mathbb{Q}] = \frac{[\mathbb{Q}(\sqrt[n/g]{a}) : \mathbb{Q}]}{[\mathbb{Q}(\sqrt[n/g]{a}) : F]} = \frac{n}{g\cdot [\mathbb{Q}(\sqrt[n/g]{a}) : F]} $$ and the modulus of the constant term of minimal polynomial of $\sqrt[n/g]{a}$ over $F$ is$$\sqrt[n/g]{a}^{[\mathbb{Q}(\sqrt[n/g]{a}):F]}=\sqrt[d]{a}.$$This is again a real field, thus $\sqrt[d]{a}\en F$. We have$$\mathbb{Q}(\sqrt[d]{a})\subseteq F \text{ and }[\mathbb{Q}(\sqrt[d]{a}): \mathbb{Q}]=d=[F:\mathbb{Q}],$$hence$$F=\mathbb{Q}(\sqrt[d]{a}).$$
Si tenemos en cuenta su lugar en el campo de $\mathbb{Q}(\zeta_{2n}\sqrt[n]{n})$ y el polinomio$$x^n+n=p_n(x),$$then the first part of the above argument again gives us$$[\mathbb{Q}(\zeta_{2n}\sqrt[n]{n}):\mathbb{Q}]\ge {n\over{g}};$$however, this time there does not have to be an equality. Together, we get $${n\over{g}} \le [\mathbb{Q}(\zeta_{2n}\sqrt[n]{n}):\mathbb{Q}]<n,$$so $g>1$.
Si $g$ es divisible por algún extraño primo, entonces hemos terminado. Ahora, suponga que $g$ es una potencia de $2$, decir $g=2^\ell$$\ell\in\mathbb{N}$. A continuación,$2\mid n$, lo $2\mid a$, so$$2^{2^\ell}=2^g\mid n.$$We have$$2^\ell\ge\ell+1,\text{ }\ell\in\mathbb{N},$$hence$$2 \text{ divides } {{n}\over{2^\ell}}={n\over{g}}.$$It follows the field$$\mathbb{Q}(\sqrt[n]{n})=\mathbb{Q}(\sqrt[n/g]{a})$$has a subfield $\mathbb{Q}(\sqrt{a})$. The fields $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{a})$ are the only subfields of $\mathbb{Q}(\sqrt[n/g]{a})$ that are Galois over $\mathbb{Q}$ because $\sqrt[d]{a}$ has non-real conjugates for $d>2$. We have $$\sqrt{2}\in\mathbb{Q}(\zeta_8)\subseteq\mathbb{Q}(\zeta_{2^{2^\ell+1}})\subseteq\mathbb{Q}(\zeta_{2n})$$ and$$\sqrt{p}\in\mathbb{Q}(\zeta_{4p})\subseteq\mathbb{Q}(\zeta_{2n})$$for every odd prime $p$ dividing $a$, hence $\mathbb{Q}(\sqrt{a})\subseteq\mathbb{Q}(\zeta_{2n})$. The extension $\mathbb{Q}(\zeta_{2n})/\mathbb{Q}$ is Galois with abelian Galois group, so $$\operatorname{Gal}(\mathbb{Q}(\zeta_{2n})/\mathbb{Q}(\zeta_{2n})\cap\mathbb{Q}(\sqrt[n]{n}))\unlhd\operatorname{Gal}(\mathbb{Q}(\zeta_{2n})/\mathbb{Q})$$and the extension $\mathbb{Q}(\zeta_{2n})\cap\mathbb{Q}(\sqrt[n]{n})/\mathbb{Q}$ is Galois (and abelian). Together, it implies$$\mathbb{Q}(\zeta_{2n})\cap\mathbb{Q}(\sqrt[n]{n})=\mathbb{Q}(\sqrt{a}).$$Since $\mathbb{Q}(\zeta_{2n})/\mathbb{Q}$ is Galois, the fields $\mathbb{Q}(\zeta_{2n})$ and $\mathbb{Q}(\sqrt[n]{n})$ are linearly disjoint over $\mathbb{Q}(\zeta_{2n})\cap\mathbb{Q}(\sqrt[n]{n})$ and$$[\mathbb{Q}(\zeta_{2n},\sqrt[n]{n}):\mathbb{Q}(\zeta_{2n})\cap\mathbb{Q}(\sqrt[n]{n})]=[\mathbb{Q}(\zeta_{2n}):\mathbb{Q}(\zeta_{2n})\cap\mathbb{Q}(\sqrt[n]{n})]\cdot[\mathbb{Q}(\sqrt[n]{n}):\mathbb{Q}(\zeta_{2n})\cap\mathbb{Q}(\sqrt[n]{n})].$$Con esto, obtenemos
$$[\mathbb{Q}(\zeta_{2n},\sqrt[n]{n}):\mathbb{Q}]$$$$=[\mathbb{Q}(\zeta_{2n},\sqrt[n]{n}):\mathbb{Q}(\zeta_{2n})\cap\mathbb{Q}(\sqrt[n]{n})]\cdot[\mathbb{Q}(\zeta_{2n})\cap\mathbb{Q}(\sqrt[n]{n}):\mathbb{Q}]
$$$$=[\mathbb{Q}(\zeta_{2n}):\mathbb{Q}(\zeta_{2n})\cap\mathbb{Q}(\sqrt[n]{n})]\cdot[\mathbb{Q}(\sqrt[n]{n}):\mathbb{Q}(\zeta_{2n})\cap\mathbb{Q}(\sqrt[n]{n})]\cdot[\mathbb{Q}(\zeta_{2n})\cap\mathbb{Q}(\sqrt[n]{n}):\mathbb{Q}]$$$$
={{[\mathbb{Q}(\zeta_{2n}):\mathbb{Q}]\cdot[\mathbb{Q}(\sqrt[n]{n}):\mathbb{Q}]}\over{[\mathbb{Q}(\zeta_{2n})\cap\mathbb{Q}(\sqrt[n]{n}):\mathbb{Q}]}}$$$$
={{[\mathbb{Q}(\zeta_{2n}):\mathbb{Q}]\cdot[\mathbb{Q}(\sqrt[n]{n}):\mathbb{Q}]}\over{[\mathbb{Q}(\sqrt{a}):\mathbb{Q}]}}$$$$={{\varphi(2n)\cdot n}\over{2^{\ell+1}}}.$$
Pero nosotros have$$\mathbb{Q}(\zeta_{2n},\sqrt[n]{n})=\mathbb{Q}(\zeta_{2n},\zeta_{2n}\sqrt[n]{n}),$$and by the same method as above, we can show that$$[\mathbb{Q}(\zeta_{2n},\zeta_{2n}\sqrt[n]{n}):\mathbb{Q}]={{[\mathbb{Q}(\zeta_{2n}):\mathbb{Q}]\cdot[\mathbb{Q}(\zeta_{2n}\sqrt[n]{n}):\mathbb{Q}]}\over{[\mathbb{Q}(\zeta_{2n})\cap\mathbb{Q}(\zeta_{2n}\sqrt[n]{n}):\mathbb{Q}]}}$$$$={{\varphi(2n)\cdot[\mathbb{Q}(\zeta_{2n}\sqrt[n]{n}):\mathbb{Q}]}\over{[\mathbb{Q}(\zeta_{2n})\cap\mathbb{Q}(\zeta_{2n}\sqrt[n]{n}):\mathbb{Q}]}}.$$Comparing these two expressions gives us$$[\mathbb{Q}(\zeta_{2n}\sqrt[n]{n}):\mathbb{Q}]={{[\mathbb{Q}(\zeta_{2n})\cap\mathbb{Q}(\zeta_{2n}\sqrt[n]{n}):\mathbb{Q}]\cdot n}\over{2^{\ell+1}}}.$$Polynomial $p_n(x)$ being reducible is thus equivalent to$$[\mathbb{Q}(\zeta_{2n})\cap\mathbb{Q}(\zeta_{2n}\sqrt[n]{n}):\mathbb{Q}]<2^{\ell+1}.$$Since $$2^\ell+1\ge\ell+2,$$we have$$\mathbb{Q}(\zeta_{2^{\ell+1}})\subset\mathbb{Q}(\zeta_{2^{\ell+2}})\subseteq\mathbb{Q}(\zeta_{2n})$$and also $\mathbb{Q}(\zeta_{2^{\ell+1}})\subseteq\mathbb{Q}(\zeta_{2n}\sqrt[n]{n})$ because $$\zeta_{2^{\ell+1}}={{(\zeta_{2n}\sqrt[n]{n})^{n\over{2^\ell}}}\over{a}}.$$Next, we have $\zeta_{2^{\ell+2}},\sqrt{a}\in\mathbb{Q}(\zeta_{2n})$ as explained above, and therefor $\zeta_{2^{\ell+2}}\sqrt{a}\in\mathbb{Q}(\zeta_{2n})$. Using once again $$2^\ell+1\ge\ell+2,$$it is also$$\zeta_{2^{\ell+2}}\sqrt{a}=(\zeta_{2n}\sqrt[n]{n})^{2n/2^{\ell+2}}\in\mathbb{Q}(\zeta_{2n}\sqrt[n]{n}).$$If we had $\zeta_{2^{\ell+2}}\sqrt{a}\notin\mathbb{Q}(\zeta_{2^{\ell+1}})$, then it would be$$[\mathbb{Q}(\zeta_{2^{\ell+1}},\zeta_{2^{\ell+2}}\sqrt{a}):\mathbb{Q}(\zeta_{2^{\ell+1}})]\ge2,$$y así
$$[\mathbb{Q}(\zeta_{2n})\cap\mathbb{Q}(\zeta_{2n}\sqrt[n]{n}):\mathbb{Q}]\ge[\mathbb{Q}(\zeta_{2^{\ell+1}},\zeta_{2^{\ell+2}}\sqrt{a}):\mathbb{Q}]$$$$=[\mathbb{Q}(\zeta_{2^{\ell+1}},\zeta_{2^{\ell+2}}\sqrt{a}):\mathbb{Q}(\zeta_{2^{\ell+1}})]\cdot[\mathbb{Q}(\zeta_{2^{\ell+1}}):\mathbb{Q}]$$$$\ge2\varphi({2^{\ell+1}})$$$$=2^{\ell+1},$$
una contradicción. Es follows$$\zeta_{2^{\ell+2}}\sqrt{a}\in\mathbb{Q}(\zeta_{2^{\ell+1}})\subset\mathbb{Q}(\zeta_{2^{\ell+2}}),$$hence $\mathbb{Q}(\sqrt{a})\subseteq\mathbb{Q}(\zeta_{2^{\ell+2}})$. But the only real quadratic subfield of $\mathbb{Q}(\zeta_{2^k})$ for $k\ge3$ is $\mathbb{Q}(\sqrt{2})$ (it also has non-real quadratic subfields $\mathbb{Q}(i)$ and $\mathbb{Q}(\sqrt{-2})$), and therefore, $\mathbb{Q}(\sqrt{a})=\mathbb{Q}(\sqrt{2})$, which by Kummer theory implies $a=2b^2$ for some $b\in\mathbb{N}$. If we had $\ell\ge2$, then it would be $$\sqrt{a}=b\sqrt{2}\in\mathbb{Q}(\zeta_8)\subseteq\mathbb{Q}(\zeta_{2^{\ell+1}}),$$and so$$\zeta_{2^{\ell+2}}={{\zeta_{2^{\ell+2}}\sqrt{a}}\over{\sqrt{a}}} \in\mathbb{Q}(\zeta_{2^{\ell+1}}),$$ which is impossible by degree comparison. Hence, finally $$\ell=1 \implies n=a^{2^\ell}=4b^4.\tag*{$\square$}$$
Un par de comentarios están en orden. En el caso de $n=4b^4$ $g=2$ (es decir, $n$ no es un $q$th potencia para cualquier extraño prime $q$ dividiendo $b$) nos have$$[\mathbb{Q}(\zeta_{2n})\cap\mathbb{Q}(\zeta_{2n}\sqrt[n]{n}):\mathbb{Q}]\ge[\mathbb{Q}(i):\mathbb{Q}]=2,$$hence$$[\mathbb{Q}(\zeta_{2n}\sqrt[n]{n}):\mathbb{Q}]={{[\mathbb{Q}(\zeta_{2n})\cap\mathbb{Q}(\zeta_{2n}\sqrt[n]{n}):\mathbb{Q}]\cdot n}\over4} \ge {n\over2}.$$This implies that the factors given by Sophie Germain's Identity are in this case irreducible over $\mathbb{Z}$ (because both factors have roots of the form $\zeta_{2n}\sqrt[n]{n}$ for some choice of primitive $2n$th root of unity $\zeta_{2n}$ since every primive $8$th root of unity is a $b^4$th power of some primitive $2n$th root of unity). However, in general, finding the factorization of $x^n\pm$ n en factores irreducibles parece difícil.
P. S. Mirando adelante a verle en un par de días!
