<s>[INST] Integer solution $(x,y,z)$ to $xyz=24$ 
Find the total number of integer solutions $(x,y,z)\in\mathbb{Z}$ of the equation $xyz=24$.

I have tried $xyz = 2^3 \cdot 3$
My Process:
Factor $x$, $y$, and $z$ as
$$
\begin{cases}
x = 2^{x_1} \cdot 3^{y_1}\\
y = 2^{x_2} \cdot 3^{y_2}\\
z = 2^{x_3} \cdot 3^{y_3}
\end{cases}
$$
Now $2^{x_1+x_2+x_3} \cdot 3^{y_1+y_2+y_3} = 2^3 \cdot 3$, so
$$
\begin{cases}
x_1+x_2+x_3 = 3\\
y_1+y_2+y_3 = 1
\end{cases}
$$
where $x_1,x_2,x_3,y_1,y_2,y_3\geq 0$. Thus the number of solutions for the first is $\binom{5}{2}=10$ and second is $\binom{3}{2}=3$. This gives a total number of positive solutions of $10 \cdot 3 = 30$
How can I calculate the total number of solutions? Is this process correct?
 [/INST] Your logic is correct for positive solutions. Now you need the sign pattern to be $+++,+--,-+-,--+$, four possibilities, so multiply by four.
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