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Find the number of solutions of $x_1+x_2-x_3-x_4=0$ in integers between $-4$ and $4$, inclusive.

I transformed $x_i$ to $y_1$ using $x_i + 4$ and my equation became $y_1 + y_2 +y_3 + y_4 = 16$ for $[0,8]$. The negative $x$ values of $x_3$ and $x_4$ are what are confusing me.

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    $\begingroup$ Please use MathJax in future :) $\endgroup$ – Shaun Nov 26 '18 at 3:59
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The number of solutions of $$x_1+x_2-x_3-x_4=0$$ where $x_i \in [-4,4]$ is equal to the number of solutions of $$z_1+z_2+z_3+z_4=0$$ where $z_i \in [-4,4]$.

To see it, suppose $(x_1,x_2,x_3,x_4)$ is a valid solution for the first system then $(x_1,x_2, -x_3, -x_4)$ is a valid solution for the second system and such mapping is a bijection.

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