no three points in a line on $\mathbb{Z}_p^2$

Question

in finite geometry, the $\mathbb{Z}_p$-plane is $\mathbb{Z}_p^2$, I have proved there are exactly $p+1$ lines pass through a given point in $\mathbb{Z}_p^2$, and by this conclusion there are at most $p+2$ points in $\mathbb{Z}_p$-plane so that no three are in a line.

so if there are exactly $p+1$ points to ensure that no three are in a line? how to construct such $p+1$ points. I've tried $p=3,5$ but the pattern is not so obvious.

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Edit Again: I've take a hard time to prove the conclusion of Lord's. additionally, if $a$ is a quadratic non-residue, $x^2-ay^2=k$ have only $1$ solution in $\mathbb{Z}_p^2$ when $k=0$, and $p+1$ solutions in $\mathbb{Z}_p^2$ when $k\ne0$. But if $a$ is a quadratic residue of module $p$, equation $x^2-ay^2=k$ will have $2p-1$ solutions in $\mathbb{Z}_p^2$ when $k=0$, and $p-1$ solutions when $k\ne 0$.

As mercio's comment, there is at most $p+2$ points that may satisfies the requirements, how to disprove there is no $p+2$ points being possible.

Answer 1

In finite geometry, a maximal set containing no three collinear points is sometimes called an oval.

In the affine plane, the solutions of a quadratic equation in two variables (a conic) has no three collinear points, as long as that equation is irreducible.

An irreducible conic in the projective plane has $p+1$ points there, so we want such a conic disjoint from the "line at infinity". If $a$ is a quadratic non-residue modulo the odd prime $p$, then a conic with equation $x^2-ay^2=1$ has no points "at infinity" and is irreducible.

I'll leave you to think about the case $p=2....$

Answer 2

If you have a set $\mathcal{O}$ of $p+2$ points in $\mathrm{AG}(2,p)$ with no three collinear, then take a point $\mathbf{x} \in \mathcal{O}$. For every other point $\mathbf{y} \in \mathcal{O}$, there is a line containing $\mathbf{x}$ and $\mathbf{y}$ but no other point of $\mathcal{O}$. This gives a total of $p+1$ lines through $\mathbf{x}$, which is all of them. So every line that meets a point of $\mathcal{O}$ actually meets two points of $\mathcal{O}$.

Okay, now take a point $\mathbf{x} \not\in \mathcal{O}$. There are some number $a$ of lines through $\mathbf{x}$ that are disjoint to $\mathcal{O}$, and some number $b$ of lines through $\mathbf{x}$ that meet $\mathcal{O}$ in two points. Can you calculate $b$? What happens if $p$ is odd?