$\mathbb{A} = \{(a,b) | a,b \in \mathbb{Z} \; and \; |a|+|b| \leq n\}$

Question

$\mathbb{A} = \{(a,b) | a,b \in \mathbb{Z} \; and \; |a|+|b| \leq n\}$
Find

• $|\mathbb{A}|$
• the number of lines passing through at least $2$ points of $\mathbb{A}$.
• the number of triangles whose vertices are points in $\mathbb{A}$

There are 2 versions. for the easy version $n = 2$.
Then we can find $|\mathbb{A}|$ in hand. There are $13$ such points. But getting messed up with other two even in easy version.
I know this approach if I am asked to find the solutions of $a + b \leq n$ -
Write the statement like this $a+b+c =n$, where $0\leq c\leq n$. Then the solution to this will be the solutions to $a + b \leq n$.
But if $|a| + |b| \leq n$ then I can't just multiply the previous solution by $4$ as there can be $(a,b), (-a,b), (a,-b), (-a,-b)$; as this will count some point multiple times, possibly only points on $x=0,y=0$ lines I guess. I dont know how to cope with them.
I will be glad to have the full solution. Any partial solution will be helpful.

This problem exists here Number of lines determined given a set of points But only with $n=2$. But I think this can be solved for any $n$.

Answer 1

The formula should be $a_n=2(n)(n+1)+1,$ where $|A|=a_n.$

Hint: Because there are $4n$ points on the boundary and we the previous lattice points contribute as well so we have $a_n=a_{n-1}+4n, $ where $a_0=1.$

For the remaining parts, I guess we could have $$\text{Number of Lines}= \binom{a_n}{2}$$ And $$\text{Number of Triangles}=\binom{a_n}{3}.$$