Roots of $x^{30}-1$ in a finite polynomial quotent field. Let $F=\mathbb{Z}/5\mathbb{Z} [x]$ (polynomials with coefficients in $\mathbb{Z}/5\mathbb{Z}$) and $I=(x^2 + \overline{2})$ be the ideal generated by $x^2 + \overline{2}$ . Consider the quotent field $F/I$. Find the elements $u \in F/I$ such that $u^{30} =1$. This is equivalent to finding the roots of $x^{30} -1$ in this strange field.
As the element $u\in F/I$ has the form $u=[\overline{a}x+\overline{b}]$ I have tried expanding $u^{30}$ using the binomial formula, then use the fact that $[x^2]=[\overline{-2}]$ to reduce the candidates and finally just use try and error.
Is there a more "elegant" way to solve this question? (With "elegant" I mean not doing so many computations).
I hope that the notation is clear.
 A: Note that
$$
x^{30}-1=(x^2 + 4x + 1)^5(x^2 + x + 1)^5(x + 4)^5(x + 1)^5
$$
in $\Bbb F_5[x]$ by Jyrki's comment. Now we can consider this modulo $I=(x^2+2)$.
A: First of all, as $5=0$ in $\mathbb{Z}/5\mathbb{Z}[x]$, we have that $(a+b)^5=a^5+b^5$ for all $a,b\in \mathbb{Z}/5\mathbb{Z}[x] $, so we get that 
$$x^{30}-1=(x^6-1)^5=((x^3-1)(x^3+1))^5=((x-1)(x+1)(x^2-x+1)(x^2+x+1))^5$$
already in $\mathbb{Z}/5\mathbb{Z}[x]$, hence in the field $K:=F/I$, which contain $\mathbb{Z}/5\mathbb{Z}$. 
Hence you only need to find the roots of each polynomial: they are $u=\pm 1$ plus the roots of $x^2\pm x+1$. A simple computation shows that 
$(\pm \bar x + 2)^2=2\mp \bar x -1$, hence $\pm \bar x+2$ is a root of $y^2+y+1$. Similarly for $(\pm \bar x + 3)$ and $y^2-y+1$. Hence the answer is $$\{\pm 1,\pm \bar x + 3,\pm \bar x + 2\}$$
A: $K=F/I$ is a field with $25$ elements. Therefore, its multiplicative group $K^\times$ is cyclic of order $24$. Finally, $u^{30}=1$ iff $u^6=u^{\gcd(30,24)}=1$. Since $K^\times$ is cyclic, it all reduces to finding a generator $g$ and taking the subgroup generated by $g^4$.
