MHB How Many Integer Solutions Exist for the Equation |x|+|y|+|z|=2010?

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The discussion focuses on the problem of finding the number of ordered triples of integers (x, y, z) that satisfy the equation |x| + |y| + |z| = 2010. The director of the Problem of the Week (POTW) acknowledges a delay in posting due to personal challenges but emphasizes the importance of community engagement in solving the problem. Members are encouraged to contribute their solutions, and there is a note of appreciation for a previous contributor's correct solution. The director expresses hope for participation and collaboration among members. Engaging with this mathematical challenge can foster problem-solving skills and community interaction.
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Here is this week's POTW:

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How many ordered triples of integers $(x,\,y,\,z)$ satisfy the equation $|x|+|y|+|z|=2010$?

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Hi MHB!

As usual, I will extend the period of solving last week's POTW until next week! I hope members will try to take a stab at the problem again and I am looking forward to receiving members' solution!
 
Hello MHB!

I want to apologize for I am late for a day to perform my duty as a director of Problem of the Week. I was (and still am) swamped with lots of work and how the prolong pandemic affected my mental well-being, and how sleep constantly eludes me and a lot of other personal issues have put me through wringer for the past week or so. Having said all that, it is utterly important at the end of the day we have to believe that we are braver than we believe, stronger than we seem and smarter than we think! (Bigsmile)

Enough is enough for the explanation and now, let me get back to business! I want to thank to Opalg for his correct solution to last two weeks' POTW, which you can find below:
For a positive integer $n$, the surface with equation $|x|+ |y| + |z| = n$ is an octahedron, with 6 vertices, 12 edges and 8 triangular faces. The points on the octahedron with integer coordinates can be counted as follows:

there is one such point at each vertex, giving a total of $6$ points;
there are $n-1$ such points on each edge (not counting the endpoints at the vertices), giving a total of $12(n-1)$ points;
there are $\frac12(n-1)(n-2)$ such points in the interior of each face, giving a total of $4(n-1)(n-2)$ points.

So the total number of points in the lattice is $6 + 12(n-1) + 4(n-1)(n-2) = 4n^2+2$.

When $n = 2010$ that is equal to $4*2010^2 + 2 = 16\,160\,402.$
 
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