Triplets divisors of 1000

Since $1000 = 2^3\cdot 5^3$ , each of $a,b,c$ must have the form $\pm 2^m5^n$ for some non-negative integers $m,n$ . As multiplication is commutative and associative, this allows us to break the problem up into three pieces:

• Count the number of ways we can choose the exponents of $2$ .
• Count the number of ways we can choose the exponents of $5$ .
• Count the number of ways we can choose the signs.

Consider just the exponents of $2$ first. If $A,B,C$ are the exponents of $2$ in $a,b,c$ , respectively then we must have $A+B+C = 3$ . Since $A,B,C$ are non-negative integers, we can count the number of ways to choose these exponents using "stars-and-bars". This gives $\binom{3+2}{2} = 10$ ways to determine the exponents of $2$ .

The exponents of $5$ are determined in the same way, giving $\binom{3+2}{2} = 10$ ways to determine the exponents of $5$ .

Then, since the product is positive, there must be an even number of negatives. Therefore, either all are positive ( $1$ way) or exactly two are negative ( $3$ ways). Therefore, there are $1+3 = 4$ ways to choose the signs.

Finally, just multiply: $\binom{3+2}{2} \cdot \binom{3+2}{2} \cdot 4 = \boxed{400}.$

Start by looking for positive solutions. There are as many choices for $a$ as there are positive divisors of $1000$ . For a given choice of $a$ , there are as many choices of $b$ as there are positive divisors of $\tfrac{1000}{a}$ . Once $a,b$ are chosen, $c$ is fixed. Thus the number of positive integer solutions of this equation is the Dirichlet convolution $(1 \star \sigma_0)(1000) \; = \; \sum_{a|1000} \sigma_0\big(\tfrac{1000}{a}\big)$ where $\sigma_0(n)$ is the number of divisors of $n$ . But $1 \star \sigma_0$ is multiplicative, and $(1 \star \sigma_0)(p^3) \; = \; \sigma_0(1) + \sigma_0(p) + \sigma_0(p^2) + \sigma_0(p^3) \; =\; 1 + 2 + 3 + 4 \; = \; 10$ for any prime $p$ , so that $(1 \star \sigma_0)(1000) \; = \; (1 \star \sigma_0)(8) (1 \star \sigma_0)(125) \; = \; 10^2 \; = \; 100$ Now we have the positive integer solutions, we find all solutions by noting that any positive integer solution could be converted into an integer solution by making any two of the three numbers $a,b,c$ negative. Thus there are an additional $3$ integer solutions for each positive integer solution, making a total of $4 \times 100 = \boxed{400}$ solutions.