Find all solutions to $xyz=1,000,000;$ $x,y,z \in \mathbb Z$

Question

I had previously solved $xy=1,000,000;$ $x,y \in \mathbb Z$, I believe: 1,000,000 has 49 factors so there are 49 pairs since $x$ and $y$ could be both positive or both negative.

Please would you help me find as efficient and systematic a way as possible.

EDIT: since there are many solutions, finding how many there are would be fine. However, unfortunately, (1,1,1000000) is considered distinct from (1000000,1,1).

Thank you

Answer 1

If $x$, $y$, and $z$ are positive integers with $xyz=2^6\cdot5^6$, then $x=2^a\cdot5^{a'}$, $y=2^b\cdot5^{b'}$, and $z=2^c\cdot5^{c'}$ with $0\le a,a',b,b',c,c'$ and $a+b+c=a'+b'+c'=6$. There are $1+2+3+4+5+6+7=28$ non-negative solutions to $a+b+c=6$, and likewise for $a'+b'+c'=6$, so there is a total of $28\cdot28$ positive triples with product $1{,}000{,}000$. Each of these triples can be modified in three different ways to give a triple with two negative signs, so the total number of integer solutions to $xyz=1{,}000{,}000$ is $4\cdot28\cdot28=3136$.

To explain the $1+2+3+4+5+6+7$, if $a=6$, then there's only $1$ choice for $b$ and $c$, namely $b=c=0$, while if $a=5$, there are $2$ choices, down to $a=0$, for which there are $7$ solutions to $b+c=6$.

Answer 2

Let $x$ run through the factors of $1,000,000$. For each $x$, find all pairs $(y,z)$ that multiply to $\frac{1,000,000}{x}$.

Answer 3

$xyz = 10^6 = 2^6 5^6 = 2 \times2 \times2 \times2 \times2 \times2 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5$

Now find the number of possible ways of dividing those 12 elements into 3 sets ($x$, $y$ and $z$).

Answer 4

hint

The only prime factors of $10^6$ are $2$ and $5$. So each of $x,y,z$ (if they are $> 1$) should have the same prime factors. Thus we have $$2^{x_1}\, 5^{x_2} \, 2^{y_1}\, 5^{y_2}\, 2^{z_1}\, 5^{z_2} = 2^{6}\, 5^{6}$$ So you need to count the number of non-negative integer triplets $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ such that \begin{align*} x_1 + y_1 +z_1 & =6\\ x_2+y_2+z_2&=6 \end{align*}

The counting for such equations can be done using $\binom{n+r-1}{r-1}$ stars and bars method.

Once you get the count of positive integer solutions, negatives will be very easy to count.