Integer solution $(x,y,z)$ to $xyz=24$

Question

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?

The election of auxiliary variables isn't precisely stellar... too easy to get lost among $x$, $x_i$ and so on jumbled together. — vonbrand, Feb 19 '13 at 02:10

score 11 · Accepted Answer · answered Feb 18 '13 at 17:23

11

Your logic is correct for positive solutions. Now you need the sign pattern to be $+++,+--,-+-,--+$, four possibilities, so multiply by four.

answered Feb 18 '13 at 17:23

Ross Millikan

374,822

Thanks Ross Millikan Got it – juantheron Feb 18 '13 at 17:34
@RossMillikan Can you help me with how OP got $5C2$ solutions for first or the one for second? It's not that I am not convinced but I am trying to learn. – Feb 18 '13 at 18:03
1

@user1709828: He was finding the number of compositions of 5 into 3 parts and of 3 into 3 parts. How to compute these is shown in the Wikipedia article. – Ross Millikan Feb 18 '13 at 18:19

score 5 · Answer 2 · answered Feb 18 '13 at 17:22

Yes, this looks right for finding the number of positive integer solutions.

Since you originally asked for all integer solutions, you'll also have to take into account the signs of the factors. There are $\binom{3}{0} + \binom{3}{2} = 4$ ways of adding zero or two minus signs, for a total of 120 integer solutions.

Integer solution $(x,y,z)$ to $xyz=24$

2 Answers2