I am working in the finite affine plane over $\mathbb F_q$ with $q=2^n$. Such a plane has $q^2$ points, $q^2+q$ lines, each line has $q$ points, and by a point is passing $q+1$ lines. There are $q+1$ directions and $q$ lines in each direction.
Here is a drawing of the plane for $q=4$ with the 5 lines going trough the point $(0,0)$. We use that $\mathbb F_4=\{0,1,x,x^2\}$ and that $x^2=x+1$ for the computation. (Don't take care of the first coordinate...)
I would like to know if there exists a set of $q+2$ points $\{P_1,...,P_{q+2}\}$ such that :
No three of them are on the same line
There are 2 by 2 equals to the same vector, i.e. $\overrightarrow{P_iP_{i+1}}=\overrightarrow{P_1P_2}$ for any odd $i$.
I know that the answer is no for $q=4$. Indeed, if such a set exists, you can without loss of generality assume that $P_1=(0,0)$, $P_2=(0,1)$, $P_3=(1,0)$ and $P_4=(1,1)$. But then all the other points are in some line $(P_iP_j)$.
I believe (and would be happy) that such a set never exists but I cant prove it. If it does not exists, what it the maximal number of points I can have with these properties ?
It is linked to the hyperoval notion in projective plane : indeed, if you add a ''infinite line" with $q+1$ points, each of them corresponding to a direction and you join the lines of the same direction to the corresponding point, you obtain $PG(2,q)$. A set of $q+2$ points with no three points on the same line in $PG(2,q)$ is an hyperoval and exists, but I don't know if the points can be 2 by 2 equals to the same vector...
Any help is welcome !