Solving an $x^2 +x-1$ in a field with $49$ elements I was given a problem that I don't seem to know how to solve. It says

Let $\mathbb{F}$ be a field with $7$ elements. Construct a field $\mathbb{L}$ with $7^2$ elements and show that $x^2-3$ and $x^2+x-1$ can be solved in $\mathbb{L}$. Now construct a field $\mathbb{K}$ with $7^3$ elements and show that $x^2-3$ cannot be solved in $\mathbb{K}$.

So far, I think that $\mathbb{F}$ must be $\mathbb{F}_7=\mathbb{Z}/7\mathbb{Z}$ and, given that $49$ is not prime, $\mathbb{L}$ must be $\mathbb{L}=\mathbb{F}_7[x]/(f(x))$ but I am having trouble finding $f(x)$. I was thinking that $f(x)=x^2-5$ would work since $\sqrt{5}\notin\mathbb{F}_7$ and $\sqrt{5}$ is needed to solve both equations. However, if I use $x^2-5$, I can't seem to solve for $\sqrt{5}$. Also, I have no idea what to do for $\mathbb{K}$.
Thanks
 A: You can take $f(x)=x^2-5$ for your field $L$.  Elements of $L$ can be considered as polynomials
$$\{a+bx\mid a,b\in F_7\}\ ,$$
with addition and multiplication modulo $x^2-5$.  To solve $t^2-3$ (maybe the $x$ is confusing you so I have changed the variable) you need
$$(a+bx)^2-3=0\ ,$$
that is,
$$a^2+2abx+b^2x^2-3=0\ .$$
since $x^2$ is "the same as" $5$ this is
$$(a^2+5b^2-3)+2abx=0\ .$$
It's fairly clear that $b\ne0$ (if it's not clear just try it) and so
$$a=0\ ,\quad 5b^2=3$$
which gives $b=\pm3$.  So the roots of $t^2-3$ are $t=\pm3x$.
You can do $t^2+t-1$ in the same way.  For $K$ you need $f(x)$ to be a cubic.
Good luck!
A: the multiplicative groups of the fields with $7^2$ and $7^3$ elements have, respectively, orders 48 and 342. thus any nonzero element present in both fields must satisfy $x^6=1$. the elements of $F_7$ already furnish all the solutions.
thus if $x^2-3$ has solutions in each field, they must be distinct. but this is not possible - otherwise the $2^{\text{nd}}$-degree equation would have four solutions in the compositum of the two fields
