Number of solutions of $|x|+|y|+|z|=10$ where $x,y,z\in\mathbb{Z}$

Question

I am trying to find the number of solutions to the integral equation $|x|+|y|+|z|=10$ where $x,y,z\in\mathbb{Z}$. Here's what I've done so far.

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My Attempt:

Let us fix a value of $|x|=k$, so $|y|+|z|=n-k$, (considering the general case of when $|x|+|y|+|z|=n$). So for each case $|y|+|z|=n-k$ has $(n-k+1)$ solutions in $|y|$ and $|z|$, and in total $4(n-k+1)$ because there's $4$ arrangements of $y$ and $z$ being $-$ or $+$, which is itself in correspondence with $2$ possibilities of $x$ when $|x|=k$, so total number of solutions should be $8\sum_{k=0}^{n}(n-k+1)=8\sum_{k=0}^{n}(k+1)=4(n+1)(n+2)$, so in this case number of solutions should be $528=48\cdot 11$.

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The actual answer is given to be $402$. Any hints are appreciated. Thanks.

Answer 1

Hint:

Note that $+0 = -0 = 0$, i.e, $0$ doesn't have separate positive & negative values. When you're counting the number of solutions for the case $|y| + |z| = n - k$, you need to account for this (i.e., don't double count the $0$ values as being positive or negative) when you're determining the total number of solutions. I'll leave it to you to account for this to determine the appropriate updated total value.

Answer 2

It suffices to count the non-negative solutions, i.e. $x,y,z\geq 0$ because if $x$ (or $y$ or $z$) is a solution, so is $-x$. Consider $3$ cases.

Case $1$: $x,y,z$ are positive. Using the stars and bars method, the number of triples is ${9\choose 2}=36$.

Case $2$: One of $x,y,z$ is $0$, and the other two are positive. Take $x=0$, for instance, to see that $y+z=10$ has $9$ positive solutions. So we have $9\cdot 3=27$ triples in this case.

Case $3$: One of $x,y,z$ is $10$, and the others are $0$. There are $3$ such triples.

In each triple from case $1$, any of the numbers can be taken with a positive or negative sign. In the triples from case $2$, this applies to exactly $2$ of the numbers (since the third one is $0$), and in the triples from case $3$, this applies only to the non-zero number. Hence, the total count is $$2^3\cdot 36+2^2\cdot 27+2^1\cdot 3=402 $$