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Let $p$ be a prime, then $\mathbb{F}_p$ is a finite field. $\mathbb{A}^2(\mathbb{F}_p)$ is an affine plane. Number of points in $\mathbb{A}^2(\mathbb{F}_p)$ is $p^2$.

I look at a line equation $ax+by=c$ and realise that number of distinct lines equals to number of triples $(a,b,c)$, where $\gcd(a,b,c)=1,\ a,b,c \in [0,p-1]$.

The question is: how to count the triples?

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Have you thought about using the fact that you know the number of points on all of these lines? I think it's a good place to start. – Patrick Da Silva Dec 7 '11 at 6:43
You are working in a field, so I don't understand the gcd condition at all? You get a line as long as one of $a,b$ is not zero, because then you can solve that coordinate as a function of the other. The remaining task is then to count how many $(a,b,c)$-combinations give rise to the same line. – Jyrki Lahtonen Dec 7 '11 at 6:48
I must say that Jyrki's comment makes me suspicious too, I didn't think about it enough : the gcd condition doesn't make sense since you cannot speak of divisors in $\mathbb F_p$, non-zero elements are units. You need to think about your question a little more! What is is exactly that you're looking for? Do you want to count the number of lines instead? – Patrick Da Silva Dec 7 '11 at 7:36
There is not only one line ; it would be naive to say that, since for every point (x,y) you can get it to solve a linear equation. It would make non-sense if every such linear equation would do the trick. – Patrick Da Silva Dec 7 '11 at 7:37
@PatrickDaSilva, a line can be represented in the form of a parametric equation $ x = x_0 + at, y = y_0 + bt, t \in \mathbb{F}_p $. There are $p$ lines if $a=0$, $p$ lines if $b=0$ and one line if neither $a$ nor $b$ is not equal to $0$. So there are $2p+1$ lines. Am I correct? – Sergey Filkin Dec 7 '11 at 7:37
up vote 3 down vote accepted

There are exactly $p$ lines that are parallel to the $y$-axis. Each of those is uniquely determined by the point of intersection with the $x$-axis.

All the other lines $y=mx+b$ are uniquley determined by their slope $m$ and intercept $b$. There are $p$ choices for both $m$ and $b$, so altogether we have $p^2$ lines that are not parallel to the $y$-axis.

Thus the answer is $p^2+p$.

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Another way of arriving at the same conclusion is that you can select the direction of the line in $p+1$ ways (parallel to $y$-axis, or a fixed slope). For each direction, you can partition the space $F_p\times F_p$ into $p$ parallel lines. The answer is then $(p+1)p$. – Jyrki Lahtonen Dec 7 '11 at 8:13
Jyrki, thank you very much, now it's clear for me. – Sergey Filkin Dec 7 '11 at 13:17

If you know about projective space, I think the easiest way is to count lines in the projective plane, and then subtract $1$ (for the line "at infinity", which you don't want to count).

By line-point duality, there are as many points as there are lines in $\mathbb{P}^2(\mathbb{F}_p)$ - and there are $p^2+p+1$ points in $\mathbb{P}^2(\mathbb{F}_p)$, because it is the disjoint union of the affine spaces of lesser or equal dimension. Thus there are $p^2+p$ lines in $\mathbb{A}^2(\mathbb{F}_p)$.

(The line-point duality comes from the fact that a line in $\mathbb{P}^2$ is given by a homogeneous equation $aX+bY+cZ=0$. It corresponds to the point $[a,b,c]$ in homogeneous coordinates.)

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Bruno, Thanks for your insightful answer. When I become acquainted with proj space I will remember this. – Sergey Filkin Dec 7 '11 at 13:15
@SergeyFilkin, you are most welcome! :) – Bruno Joyal Dec 7 '11 at 15:45

I think what you want to do is to count the number of distinct lines. Your condition with the GCD only works when you are in $\mathbb Z$ or something like that. In $\mathbb F_p$ it's a little different. In the following I consider two lines to be the same if they have the same set of solutions, just to make sure I am clear.

There are $p$ lines of the form $ax = c$ and $a \neq 0$, because the set of solutions for those is just $(a^{-1} c, y)$ with $y$ ranging from $0$ to $p-1$ ; hence $a^{-1}c = A$ for some $A$ and the lines are just $(A,y)$ with $y \in \mathbb F_p$. There is also $p$ lines of the form $by = c$ with $b \neq 0$ for the same reasons. If $a=b=c=0$ then the line is the set of all couples $(x,y)$. If $a=b=0 \neq c$ then you get the empty line. Hence by considering $a=0$ or $b=0$ I got $2p+2$ lines. You distinguish what you consider a line or not in the latter. (You might not want to consider the empty line as a line.)

Assume from now on that $a \neq 0 \neq b$. A line of the form $ax + by = 0$ is always of the form $y = Ax$ with $A = b^{-1}a$. There are $p-1$ such lines, since $A$ can range from $1$ to $p-1$ and clearly all those lines are distinct.

It remains to consider the case where $c \neq 0$. In fact, if $ax + by = c$, then $y = Ax + C$, and by translating we have $y = A(x-A^{-1} C) + C = Ax - C + C = Ax$. Hence adding the constant $c$ gives us $p-1$ new lines for each value of $A$, and they are just translates of the old lines of the form $y=Ax$.

In the end, there are $(2p+2) + p(p-1) = p^2 + p + 2$ lines if you consider the empty set as a line, and $p^2 +p+1$ if you don't.

Hope that helps,

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Sorry if I kept editing. I am thinking about the answer while I am writing it so I corrected stuff along the way. – Patrick Da Silva Dec 7 '11 at 8:11
I did everything explicitly to make sure I was clear, but the right way to think about this is the following : look at a $p \times p$ grid and forget about the cases where the line is the whole grid or the empty set. A line in there is either vertical or of the form $y=ax+b$. There are $p$ vertical lines and $p^2$ choices of $a$ and $b$, and if $y=ax+b$ has the same set of solutions than $y=Ax+B$ then $a=A$ because you can compute slopes, and $b=B$ because $a=A$. Hence you get $p^2$ lines with $y=ax+b$ and $p$ lines with $x=c$, thus getting $p^2 + p$ lines. In the end I think I got it right. – Patrick Da Silva Dec 7 '11 at 8:16
thanks alot. Now it's clear. – Sergey Filkin Dec 7 '11 at 13:46
You must have counted something twice, because the answer is $p^2+p$... also, why would the empty set ever count as a line? o_O – Bruno Joyal Dec 7 '11 at 22:22
Surely a line = a coset of a one-dimensional subspace? Ok, in projective space we need to modify that in the usual way :-) – Jyrki Lahtonen Dec 26 '11 at 17:58

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