Find all square roots of $x^2+x+4$ in $\mathbb{Z}_5[x]$. 
Find all square roots of $x^2+x+4$ in $\mathbb{Z}_5[x]$. Also show that in $\mathbb{Z}_8[x]$ there are infinitely many square roots of $1$. 

I know how to find the square roots when in $\mathbb{Z}[x]$ however I keep getting confused with $\mathbb{Z}_5[x]$. From here I am assuming I will be able to figure out exactly why the second statement is true.
 A: $5$ is small enough to do this by force:  $(ax+b)^2 \equiv x^2+x+4 \pmod{5}$ 
$$a^2 \equiv 1;\ 2ab \equiv 1;\ b^2 \equiv 4.$$  
$a^2 \equiv 1 \implies a \equiv 1 \mbox{ or } 4$
$b^2 \equiv 4 \implies b \equiv 2 \mbox{ or } 3$
Of the four combinations, $2\cdot 1 \cdot 3 \equiv 1$ and $2\cdot 4 \cdot 2 \equiv 1$ so the roots are
$$
x+3 \\
4x+2
$$
The reason these are the only roots is that if you have a higher power of $x$, such as in 
$(mx^2 + a^x + b)^2$, then the $m^2 x^4$ (or in general $m^2 x^{2k}$) higest power has a coefficient which cannot be one because the only number whose square is zero (in mod 5) is zero.
Now go to $\Bbb{Z}_8[x]$.  Here we have a way to go to higher powers in the square root, because $4^2 \equiv 0$.  In fact, if $(ax+b)^2 \equiv 1$
then for any polynomial $P(x)$,
$$(4x^2P(x)+ax+b)^2 \equiv 16(P(x)^2 x^4+ 8P(x)x^2+  1 \equiv 1$$
Since there are infinite possible polynomials, there are infinite roots if there is at least one root.  And since $1$ is a root, there are infinitely many roots.
A: Hints:
A square root of $x^2+x+4$ has the form $\pm x+a$. Let's identify:
$$(\pm x+a)^2=x^2\pm 2a x+a^2=x^2+x+4.$$
We have to solve $\;\begin{cases}a^2=4\\\pm2a=1\end{cases}$.
The first equation has solutions $\;a=\pm 2$. Checking the second equation, we find the square roots
$$x-2\enspace\text{or}\enspace -x+2. $$
For the second question, try  binomials $ax^n+b$, where the leading coefficient $a$ is nilpotent.
A: The first assertion can be found without brute force but with a little logic. if $p := x^2+x+4$ is a square then it has a double root (a root with multiplicity 2) so $p$ is divisible by it's derivative $p' = 2x+1$. Euclidean division then gives $x+3$ with, of course, remainder 0.
Let $q_1 \in \Bbb Z_8[x]$ of degree $1$  s.t. $q^2_1 = 1$ then define $q_i = 4x^i + q_{i-1}$. Now $q_i$ has degree $i$ and $q_i^2 = q_{i-1}^2 = 1$. A solution for $q_1$ is $4x+1$. 
