7. Prove that one of $2$, $3$, or $6$ is a square in $\mathbb{F}_p$. Conclude$$x^6-11x^4+36x^2-36=(x^2-2)(x^2-3)(x^2-6)$$has a root modulo $p$ for every prime $p$ but has no root in $\mathbb{Z}$.
11. Prove that $x^{p^n}-x+1$ is irreducible over $\mathbb{F}_p$ only when $n=1$ or $n=p=2$.
Proof: (6) Lemma 1: Let $f(x),g_i(x)∈\mathbb{Z}[x]$ be irreducible with finite indexing set $I$. If $θ∈\mathbb{Q}(α_1,...)$ for some roots $g_i(α_i)=0$ and the coefficients of $θ$ are given in integers, then if $Φ$ is reduction modulo $p$ and $β_i$ is a root of $Φ(g_i(x))$ we have $θ'$ (the corresponding integers having been reduced mod $p$, and $α_i$ replaced with $β_i$) is a root of $Φ(f)$. Proof: $Φ$ is an additive homomorphism $\mathbb{Z}(α_i)→\mathbb{F}_p(β_i)$ by its construction on a basis, and further it is multiplicative between two basis elements $Φ(\prod α_i)Φ(\prod α_j)=Φ(\prod α_i \prod α_j)$. Thus it is a ring homomorphism and we see $Φ(f)(θ')=Φ(f(θ))=Φ(0)=0$.$~\square$
Now, since a biquadratic extension of $\mathbb{F}_p$ would entail only the four automorphisms negating $\sqrt{D_1}$ and $\sqrt{D_2}$, none of which having order $4$ and thus none representing Frobenius $σ_p$, we must have $\mathbb{F}_p(\sqrt{D_1},\sqrt{D_2})$ is contained in $\mathbb{F}_{p^2}$ and now $m_θ(x)$ of degree $4$ cannot be irreducible mod $p$. With some work we find $\pm \sqrt{2} \pm \sqrt{3}$ are the solutions to the irreducible $x^4-10x^2+1$ and generate $\mathbb{Q}(\sqrt{2},\sqrt{3})$. Hence $m_{\sqrt{2}+\sqrt{3}}(x)=x^4-10x^2+1$ generates only a second degree extension of $\mathbb{F}_p$ and is not irreducible.
(7) Lemma 2: One of $a,b,ab∈\mathbb{F}_{p^n}$ is a square in $\mathbb{F}_{p^n}$. Proof: Since $\_^2$ is a multiplicative endomorphism of $\mathbb{F}_{p^n}^×$ and for each square $α^2$ we have only two possible elements in the fiber by $x^2-α^2=(x-α)(x+α)$, we must have $[\mathbb{F}_{p^n}^×~:~\text{img }\_^2]≤2$. Now if $a$, $b$ are not squares then $a,b \neq 1$ in $\mathbb{F}_{p^n}^×/\text{img }\_^2$. Hence $ab=1$, i.e. $ab$ is a square.$~\square$
Thus we have proved either $2$, $3$, or $6$ is a square in $\text{F}_p$ and thus the polynomial above has a root in $\text{F}_p$.
(11) We note that if $α$ is a root, then so is $α+a$ for any $a∈\mathbb{F}_{p^n}$. Thus $\mathbb{F}_{p^n}⊆\mathbb{F}_p(α)$ as else the latter would contain all the roots of $x^{p^n}-x+1$ and then a nontrivial extension would yet more roots, a contradiction. Now we observe any automorphism of $\text{Gal}(\mathbb{F}_{p}(α)/\mathbb{F}_{p^n})$ must be defined by $α↦α+a$ for some $a∈\mathbb{F}_{p^n}$ and since these automorphisms fix $\mathbb{F}_{p^n}$ they all have order $p$. Since a Galois group of degree $k$ is always cyclic over $\mathbb{F}_p$ with generator $σ_p$ and hence always cyclic over $\mathbb{F}_{p^n}$, so $[\mathbb{F}_p(α)~:~\mathbb{F}_{p^n}]=p$.
Thus we must have $pn=p^n$. We write $n=p^k$ for some $k≥0$. If $k > 1$ then we notice $k+1=p^k$ and since $p≥2$ implies $k+1 < 2^k ≤ p^k$ when $k=2$ and inductively $(k+1)+1 < 2^k+1 ≤ 2^k+(2^{k+1}-2^k) = 2^{k+1} ≤ p^{k+1}$ we must have $k=0$ (and $n=1$) or $k=1$ (in which case $n=p$ and $p^2=p^n$ and $n=p=2$). We have previously verified $x^p-x+1$ is irreducible ($γ$ implies $γ+1$ a root, thus all roots are in the same extension, thus there is the same degree $d$ among $k$ irreducibles implying $kd=p$ and $d=1$, but there are no roots in $\mathbb{F}_p$) and one may check $x^4-x+1∈\mathbb{F}_2[x]$ has no roots and is not the square of the only irreducible quadratic $(x^2+x+1)^2=x^4+x^2+1$.$~\square$
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