Find the number of integer solutions of $|x|+|y| \le 10$

Question

Find the solutions of $|x|+|y| \le 10$, where $x$ and $y$ are integers.

My solution:

$$|x|+|y|+z=10$$

Now the solutions if there were no absolute values is:

$$\binom{13-1}{10}=\frac{11*12}{2}=66$$

now subtract that once that have $0$ then multiply the others by $4$ and multiply that ones that only has $1$ zeroes by $2$.I get $221$ Am I right?

Answer 1

The number of integer solutions of $|x|+|y|\le n$ is $$1+4n+4\sum_{k=1}^{n-1} k=2n(n+1)+1.$$ $1$ is for the origin, $4n$ is for the points on the 4 semi-axis, and finally $4\sum_{k=1}^{n-1} k$ counts the points in the 4 right triangles inside the 4 quadrants.

See also the Sloane' sequence https://oeis.org/A001844

Answer 2

Pick's Theorem says that the area of a polygon whose vertices have integer coordinates is given by

$$A=I+{B\over2}-1$$

where $I$ is the number of Interior points with integer coordinates and $B$ is the number of Boundary points with integer coordinates. For the given problem the polygon is a square with diagonals of length $20$, so $A=200$. The bounday points satisfy $|x|+|y|=10$, so it's easy to see that $B=40$, hence $I=200-{40\over2}+1=181$. The number we want is

$$I+B=181+40=221$$