# Elliptic curves with trivial Mordell–Weil group over certain fields.

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?

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: $1, C_2, C_3,C_4, C_5,C_6,C_7, C_2\times C_2$.
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.