= Solution
The squares in the <finite field> $\mathbb F_7$ are $0,1,2,4$. For $x=0,1,\ldots,6$, the values of $x^3-x+1$ and the numbers of possible $y$ are respectively
$$
\begin{array}{c|rrrrrrr}
x&0&1&2&3&4&5&6\\\hline
x^3-x+1&1&1&0&4&5&2&1\\
\#y&2&2&1&2&0&2&2
\end{array}
$$
Adding $O$ yields \b[$\#E(\mathbb F_7)=12$]. The <Trace of Frobenius> is $a=8-12=-4$. The <trace of the square of an elliptic-curve endomorphism> is $a^2-2q=16-14=2$, so the <elliptic-curve point count over a finite field> gives
$$
\boxed{\#E(\mathbb F_{49})=49+1-2=48.}
$$
For example, this trace identity follows from $\pi^2-[a]\pi+[q]=0$ and $\operatorname{tr}[q]=2q$.
One suitable second <elliptic curve> is
$$
\boxed{E^{\prime}:y^2=x^3+3x+6\quad\text{over }\mathbb F_7.}
$$
Here $4\cdot3^3+27\cdot6^2\equiv2\pmod7$, so its <elliptic-curve discriminant> is nonzero. Its complete list of <rational points> is $O,(3,0),(6,3),(6,4)$, obtained by checking the seven $x$-values. For $P=(6,3)$ the tangent slope is $1$ in $\mathbb F_7$, and the <elliptic-curve addition formula> gives $2P=(3,0)$. Consequently $P$ has order four and \b[$E^{\prime}(\mathbb F_7)\cong\mathbb Z/4\mathbb Z$].
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