<s>[INST] Show that $\langle 5, x^2+x +1 \rangle$ is maximal ideal in $\mathbb{Z}[x]$. Here is my try, of which I'm rather skeptical. 
Let $I$ an ideal such that $\langle 5, x^2+x +1 \rangle \subset I \subset \mathbb{Z}[x]$. Because of the containment, there must be some $\alpha \in I$ such that:
$$\alpha \mid (5 + x^2+x+1) \implies (\alpha \mid 5) \wedge (\alpha \mid x^2+x+1).$$
Since both $5$ and $x^2 +x+1$ are irreducible, then $\alpha=1$, thus $I=\mathbb{Z}[x]$. 
Thanks!
 [/INST] Assume by contradiction
$$ \langle 5, x^2+x +1 \rangle \subsetneq I \subsetneq \mathbb{Z}[x] $$
Pick some $P(x) \in I$ which is not in $\langle 5, x^2+x +1 \rangle$. 
By long division you can write $P(x) =(X^2+X+1)Q(X)+aX+b$.
As $X^2+X+1 \in  \langle 5, x^2+x +1 \rangle \subsetneq I$ you get $aX+b \in I$ and $aX+b \notin \langle 5, x^2+x +1 \rangle$. 
Case 1: If $5|a$ then $aX \in \langle 5, x^2+x +1 \rangle$ from which you get $b \in I$ and $b \notin \langle 5, x^2+x +1 \rangle$.
The former implies that $5$ doesn't divide $b$, and hence $1$ can be written as a linear combination of $5,b$, thus $1 \in I$ (contradiction).
Case 2: If $5 \nmid a$, then there exists a $k$ such that $ak=5n+1$. Then, as $5 \in I$,
$$akx+bk \in I \implies X+bk \in I$$
By dividing $bk$ by $5$ you get that $X+c \in I$ for some $c \in \{ 0, \pm 1 , \pm 2 \}$.   
Then by long division of $X^2+X+1$ to $X+C$ the remainder is an integer which must be relatively prime to $5$. Thus again you get $1 \in I$, contradiction.
Alternately for the long division at the end, observe that:
If $X \in I$ then 
$$1 =X^2+X+1-X(X+1) \in I$$
If $X + 1 \in I$ then 
$$1=X^2+X+1-(X+1)X \in I$$
If $X - 1 \in I$ then 
$$1=2(X^2+X+1)-2(X-1)(X+2)-5\in I$$
If $X +2  \in I$ then 
$$1=2(X^2+X+1)-2(X-1)(X+2) -5\in I$$
If $X - 2 \in I$ then 
$$1=3(X^2+X+1)-3(X+3)(X-2)-20 \in I$$
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