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

My solution:


Now the solutions if there were no absolute values is:


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?

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    $\begingroup$ Do you want the solutions, or the number of solutions? $\endgroup$ – Arthur Aug 4 '16 at 11:09
  • $\begingroup$ @Arthur number of the solutions. $\endgroup$ – Taha Akbari Aug 4 '16 at 11:09
  • $\begingroup$ i would start with $$x=-10,y=0$$ or $$x=-9,y=-1$$ or $$x=-9,y=0$$ or $$x=-9,y=1$$ and so on $\endgroup$ – Dr. Sonnhard Graubner Aug 4 '16 at 11:10
  • $\begingroup$ are you interested on all solutions? $\endgroup$ – Dr. Sonnhard Graubner Aug 4 '16 at 11:14
  • $\begingroup$ There is a nice graphical approach. $|x| + |y| \le 10$ describes the interior of a square with vertices $(0,\pm10), (\pm10,0)$. Simply shade in the integer points $(x,y)$ lying within this square. $\endgroup$ – Mr. Chip Aug 4 '16 at 11:15

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


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


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



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