<s>[INST] $x^4+x^3+x^2+x+1$ irreducible over $\mathbb F_7$ This question came from the answer here.
The answer claims that $x^4+x^3+x^2+x+1$ is irreducible over $\mathbb F_7$. I can check that it has no roots in the field, but why can't it be written as a product of two quadratics? I tried the method of undetermined coefficients: $$x^4+x^3+x^2+x+1=(x^2+ax+b)(x^2+cx+d)=\\ x^4+(a+c)x^3+(ac+b+d)x^2+(bc+ad)x+bd$$ so $$a+c=1\\ac+b+d=1\\bc+ad=1\\bd=1$$ but I wasn't able to solve this. Maybe there is a way to see irreducibility by only using general finite fields theory?
 [/INST] If you want to continue your approach in case you cannot come up with some clever argument like that in Alon Amit's answer, then you can do it as follows.  First, note that $c=1-a$ and $d=b^{-1}$.  Thus, the equation $ac+b+d=1$ becomes
$$a(1-a)+b+b^{-1}=1\,.\tag{*}$$
The equation $bc+ad=1$ is now
$$b(1-a)+ab^{-1}=1\,.$$
Thus,
$$a^{-1}\big(1-b(1-a)\big)=b^{-1}=1-b-a(1-a)\,.$$
That is,
$$1-b(1-a)=a-ab-a^2(1-a)\,.$$
Consequently,
$$(2a-1)b=-1+a-a^2+a^3=(a-1)(a^2+1)\,.\tag{#}$$
Thus, $a\notin\{0,1,4\}$ for the equation above to be possible (noting that $a\neq 0$ and $b\neq 0$).  You are left with four choices of $a$, namely, $a\in\{2,3,5,6\}$.
If $a=2$, then, using (#), we get $b=3^{-1}\cdot 5=5\cdot 5=4$, so $b^{-1}=2$, but then $$a(1-a)+b+b^{-1}=2\cdot(-1)+4+2=4\neq 1\,.$$
If $a=3$, then we have by (#) that $b=5^{-1}\cdot20=-3=4$, whence
$$a(1-a)+b+b^{-1}=3\cdot(-2)+4+2=0\neq 1\,.$$
If $a=5$, then $b=3^{-1}\cdot (-1)=-5=2$, so $b^{-1}=4$, but then
$$a(1-a)+b+b^{-1}=5\cdot(-4)+2+4=0\neq 1\,.$$
Finally, if $a=6$, then $b=4^{-1}\cdot3=2\cdot 3=-1$, whence $b^{-1}=-1$, and
$$a(1-a)+b+b^{-1}=6\cdot(-5)+(-1)+(-1)=-4\neq 1\,.$$
Thus, (*) cannot be satisfied.
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