 Solution

The squares in the finite field ${\mathbb{F}}_7$ are $0,1,2,4$. For $x=0,1,\dots ,6$, the values of $x^3-x+1$ and the numbers of possible $y$ are respectivelyAdding $O$ yields $\#E({\mathbb{F}}_7)=12$. 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 givesFor example, this trace identity follows from ${\pi }^2-[a]\pi +[q]=0$ and $\mathrm{tr}[q]=2q$.

One suitable second elliptic curve isHere $4\cdot 3^3+27\cdot 6^2\equiv 2(\mathrm{mod}7)$, 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 $E^{\prime }({\mathbb{F}}_7)\cong \mathbb{Z}/4\mathbb{Z}$.