Consider the ordinary elliptic curves 
$$
E\!:y\_1^2 + x\_1y\_1 = x\_1^3 + 1,\qquad E^\prime\!: y\_2^2 + x\_2y\_2 = x\_2^3 + x\_2^2 + 1
$$
over the field $\mathbb{F}\_2$. They are quadratic twists to each other. I checked that the Kummer surface of $E \!\times\! E^\prime$, i.e., the quotient $E \!\times\! E^\prime/[-1]$ has the affine model
$$
K\!: y^2 + x\_1x\_2y = (x\_1x\_2)^2(x\_1 + x\_2 + 1) + (x\_1 + x\_2)^2,
$$
where $y := x\_1y\_2 + y\_1x\_2$.


Let $t := x\_1$, $x := x\_2$. It is almost obvious that the elliptic curve $K\_t/\mathbb{F}\_2(t)$ is reduced (over $\mathbb{F}\_2(t)$) to the form
$$
\mathcal{E}\!:y^2 + txy = x^3 + t(t^2 + t + 1)x^2 + t^4x.
$$


Is there a way to find any $\mathbb{F}\_2(t)$-point on $\mathcal{E}$ outside $$\mathcal{E}[2] = \{(0:0:1), (0:1:0)\}?$$



Is there a way to find any $\mathbb{F}_2(t)$-point on the elliptic curve $\mathcal{E}$?
Any $\mathbb F\_2(t)$-point of $K\_t$ would give, upon pullback to $E \times E'$, a $\mathbb F\_2(E)$-point of $E'$. Because $E$ is ordinary, $a\_2(E)\neq 0$, hence $E$ is not isogenous to its quadratic twist $E'$, so any such point arises from an $\mathbb F\_2$-point of $E'$. Because $E'$ has two $\mathbb F\_2$-points, these are the only ones.


