= Solution
Count affine solutions and then add the point at infinity. Over the <finite field> $\mathbb F_2$, both values of $x$ give $x^3-x=0$, and both values of $y$ solve $y^2+y=0$. Hence there are four affine points. Over $\mathbb F_3$, all three values of $x$ give zero, and each has the two ordinates $0,2$, giving six affine points.
Over the <finite field> $\mathbb F_{11}$, completing the square gives $(2y+1)^2=1+4(x^3-x)$. The square residues are $0,1,3,4,5,9$. For $x=0,1,\ldots,10$, the right sides and the corresponding numbers of ordinates are
$$
\begin{array}{c|rrrrrrrrrrr}
x&0&1&2&3&4&5&6&7&8&9&10\\\hline
1+4(x^3-x)\bmod11&1&1&3&9&10&8&5&3&4&10&1\\
\#\{y\}&2&2&2&2&0&0&2&2&2&0&2
\end{array}
$$
Their sum is sixteen. The resulting <elliptic-curve point counts over a finite field> are
$$
\boxed{\#\widetilde E_2(\mathbb F_2)=5,\qquad\#\widetilde E_3(\mathbb F_3)=7,\qquad\#\widetilde E_{11}(\mathbb F_{11})=17.}
$$
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