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I am looking for elliptic curves $E,E'$ defined over $\mathbb{F}_{3}$ and $\mathbb{F}_{4}$ respectively and given by a Weierstrass equation such that their Mordell-Weil group is trivial, i.e. such that $$ E(\mathbb{F}_{3})=\{\mathcal{O}\}\quad\text{and}\quad E'(\mathbb{F}_{4})=\{\mathcal{O}\}, $$ where $\mathcal{O}$ is the point at infinity. Is there any procedure to find them? Or is the only way just to change the coefficients until you find one satisfying the requirements?

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By Hasse's bound we know that $1\le |E(\mathbb{F}_3)|\le 7$; and indeed there is an elliptic curve with $E(\mathbb{F}_{3})=\{\mathcal{O}\}$, given by $$ y^2=x^3-x-1. $$ Actually, since we know that all such curves are given by the long Weierstrass equation $y^2=x^3+ax^2+bx+c$ with nonzero discriminant, we can just try all possibilities for $a,b,c$. There are not many curves to test. Taking all possibilities we obtain that $E(\mathbb{F}_{3})$ can be one of the following possibilities: $1, C_2, C_3,C_4, C_5,C_6,C_7, C_2\times C_2$.

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Let $K=\Bbb{F}_4=\{0,1,\alpha,\alpha+1\}$ with $\alpha^2=\alpha+1$. For all $x\in K*$ we have $x^3=1$. Therefore the cubic $x^3+\alpha$ only takes values in $K\setminus\Bbb{F}_2$. On the other hand, $y^2+y\in\Bbb{F}_2$ for all $y\in K$.

Therefore the equation $$ y^2+y=x^3+\alpha $$ has no solutions $(x,y)\in K^2$. The point at infinity is thus the only $K$-rational point of this elliptic curve.

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