Number of pairs $(i,j)$ less than or equal to $m$ Given a number $m\in\mathbb{N}$, how many pairs $(i,j)\in \mathbb{Z}^2$are there such that $$|i|+|j|\leq m?$$
For example, for $m=1$ one has $5$ pairs: $(0,0),(1,0),(0,1),(-1,0),(0,-1)$. What about in the general case?
 A: If $i>0$, $j\ge0$ and $i+j\le m$, then $(j,-i),(-i,-j),(-j,i)$ are distinct solutions; these four sets of solutions cover the entire solution space save for $(0,0)$. Thus the number of solutions is four times the number of solutions in a quadrant plus one or $4\left(\frac{m(m+1)}2\right)+1=2m^2+2m+1$.
A: Purely Enumerative Approach: Do this in two cases, either you have a $0$ or not. If you have a $0,$ there are $1+2\cdot 2\cdot m$ ways to do this (why?). If you do not have a $0,$ you have $4$ ways to choose the signs and in how many ways can you have $i+j\leq m$ for $i,j>0$? Use stars and bars or simply fix $i$ and iterate $j.$
You should get $1+4m+4\binom{m}{2}=1+4\left (\binom{m}{1}+\binom{m}{2}\right )=1+4\binom{m+1}{2}.$

For the fun (Enumerative Geometry Approach): Use Pick's theorem. Notice that $2m^2$ is the area of the square and Pick's theorem says that $2m^2=i+b/2-1,$ where $i$ are the number of reticular internal points, and $b$ the reticular boundary points. Notice further that the number of boundary points are $4m$ points. So $4m^2=2i+b-2.$ Notice that you want $i+b$ to be your answer(why?), call it $x$ so $4m^2-i+2=4m^2-x+4m+2=x$ implying the result.
A: Let
$a_n=\{(i,j)\in \mathbb Z^2: |i|+|j|\leq n\}$ the number integer-coordinate points inside the $\ell_1$ ball with radius $n$. We know that $a_0=1$.
Based on this geometric intuition, it is not hard to see that
$$a_{n+1}=a_n+4n.$$
By induction, we have
$$a_n=1+4.1+4.2+...+4.n=1+4\frac{n(n+1)}{2}=1+2n(n+1).$$
