<s>[INST] Show that $x^2+x+4$ is irreducible over $\mathbb Z_{11}$. 
Show that $x^2+x+4$ is irreducible over $\mathbb Z_{11}$. 

The problem can be solved by putting every element from $\mathbb{Z}_{11}$ and checking if there is any root.
$0^2 + 0 + 4 = 4 \\
1^2 + 1 + 4 = 6 \\
2^2 + 2 + 4 = 10 \\
3^2 + 3 + 4 = 5 \\
4^2 + 4 + 4 = 2 \\
5^2 + 5 + 4 = 1 \\
6^2 + 6 + 4 = 2 \\
7^2 + 7 + 4 = 5 \\
8^2 + 8 + 4 = 10 \\
9^2 + 9 + 4 = 6 \\
10^2 + 10 + 4 = 4 $ 
But this process is very long one. I am seeking a short method by which I can solve such problem. 
Now suppose that it is reducible, then the factors should be linear. So how can we solve the problem based on this kind of observation? Any help will be great. Thanks.
 [/INST] As we can use the formula for the roots of a quadratic over any field with characteristic $\;\neq2\;$, calculating this quadratic's discriminant we get
$$\Delta=1-4\cdot4=-15=-4=7\pmod{11}$$
and now check the squares modulo $\;11\;$ :
$$0^2=0\;,\;\;1^2=1\;,\;\;2^2=4\;,\;\;3^2=9\;,\;\;4^2=5\;,\;\;5^2=3\;,\;\;6^2=3$$
and that is enough (why?), and thus there exists no $\;x\in\Bbb F_{11}\;$ with $\;x^2=7\;$ so the quadratic has no roots in $\;\Bbb F_{11}\;$ and is thus irreducible.
We could, of course, also use quadratic reciprocity instead of calculating the squares modulo $\;11\;$ (this is highly advisable for big primes) :
$$\left(\frac7{11}\right)\stackrel{Q.R}=-\left(\frac47\right)=-1$$
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