Prove that $(x^2+x)^{2^n}+1$ is irreducible

Question

Prove that $f(x)=(x^2+x)^{2^n}+1$ is irreducible. I've thought about using mod 2 or $\mathbb Z_2$, which results in $f(x)=x^{2^{n+1}}+x^{2^n}+1=(x^{2^n}+x^{2{n-1}}+1)(x^{2^n}-x^{2{n-1}}+1)$ I've also thought about using $\mathbb Z_4$, which results in $f(x)=x^{2^{n+1}}+2x^{3*2^{n-1}}+x^{2^n}+1$, for which I'm not sure how to proceed.

EDIT by Sil: This is a problem from Romanian IMO Team Selection Tests 1998.

Answer 1

Assume that $f(x)=(x^2+x)^{2^n}+1=g(x)h(x)$ is reducible. Then it must be reducible in $\mathbb{Z}_2$. Note $f_2(x)=(x^2+x+1)^{2^n}$ is the function in $\mathbb{Z}_2$, and $x^2+x+1$ is irreducible. Define $g_2(x), h_2(x)$ similarly, and since $\mathbb{Z}_2$ is a UFD, let's just set $g_2(x)=(x^2+x+1)^y$ and $h_2(x)=(x^2+x+1)^{2^n-y}$.

Let $g(x)=g_2(x)+2g_1(x)$, $h(x)=h_2(x)+2h_1(x)$. Let $r$ be a root of $x^2+x+1$. If $g,h$ are nonconstant, then $g_2(r)=h_2(r)=0$ From $g(x)h(x)=f(x)$, plugging $x=r$, we obtain $4g_1(r)h_1(r)=(r^2+r)^{2^n}+1=2$. Note that $g_1(r)h_1(r)=ar^2+br+c$ for some integers $a,b,c$ because $r^3=1, r^4=r, r^5=r^2$. Also $r^2+r+1=0$, so WLOG $c=0$. Note that $a(-\frac{1}{2}+\frac{\sqrt{3}i}{2})+b(-\frac{1}{2}-\frac{\sqrt{3}i}{2})$ is real, so $a=b$, but then $a=-\frac{1}{2}$ is an integer. Contradiction.

Answer 2

Look at the 2-adic integers $$f(x)= (x^2+x)^{2^n}+1= \prod_{a=1}^{2^n} (x^2+x-\zeta_{2^{n+1}}^{2a+1})\in \Bbb{Q}_2[x]$$ Let $L=\Bbb{Q}_2(\beta)$ with $\beta$ a root of $x^2+x-\zeta_{2^{n+1}}$. It contains $K = \Bbb{Q}_2(\zeta_{2^{n+1}})$ the splitting field of $\Phi_{2^{n+1}}(x)=x^{2^n}+1 \in \Bbb{Q}_2[x]$.

Let $(\pi)$ be $O_K$'s maximal ideal.

From $\zeta_{2^{n+1}}^{2^{n+1}} = 1$ and that $O_K/(\pi)$ is a field with $2^m$ elements we know $\zeta_{2^{n+1}} \equiv 1 \bmod (\pi)$ and $\sum_{l=0}^{2a} \zeta_{2^{n+1}}^l \equiv 2a+1\equiv1 \bmod (\pi)$ which means

$$v_2(\zeta_{2^{n+1}}^{2a+1}-1)=v_2(\zeta_{2^{n+1}}-1)+v_2(\sum_{l=0}^{2a} \zeta_{2^{n+1}}^l)=v_2(\zeta_{2^{n+1}}-1)$$

Together with $$\Phi_{2^{n+1}}(x+1)= (x+1)^{2^n}+1=\prod_{l=1}^{2^n} (x+1-\zeta_{2^{n+1}}^{2a+1}), \qquad \prod_{l=1}^{2^n} (1-\zeta_{2^{n+1}}^{2a+1})=\Phi_{2^{n+1}}(1)=2 $$ we find $v_2(\zeta_{2^{n+1}}-1) =\frac{\sum_{l=1}^{2^n} v_2(1-\zeta_{2^{n+1}}^{2a+1})}{2^n}= \frac{v_2(2)}{2^n}=\frac1{2^n}=\frac1{\deg(\Phi_{2^{n+1}})}$.

Thus $\Phi_{2^{n+1}}$ is irreducible, $K/\Bbb{Q}_2$ is totally ramified and $\pi=1-\zeta_{2^{n+1}}$ is an uniformizer.

We obtain $$x^2+x-\zeta_{2^{n+1}} \equiv x^2+x-1 \in O_K/(\pi)[x]\cong \Bbb{Z/2Z}[x]$$ Which is irreducible, and since $L$ is $x^2+x-\zeta_{2^{n+1}}\in K[x]$ splitting field it means $$[L:K] = 2, \qquad [L:\Bbb{Q}_2]=[L:K][K:\Bbb{Q}_2] = 2^{n+1}=\deg(f)$$

And hence $f$ is irreducible over $\Bbb{Q}_2$ and it stays irreducible over $\Bbb{Q}$.