Final answer:
There are five ordered integral triples (a, b, c) where each number is the product of the other two: (0, 0, 0), (1, 1, 1), (-1, -1, 1), (-1, 1, -1), and (1, -1, -1).
Step-by-step explanation:
The question asks how many ordered integral triples (a, b, c) satisfy the condition that each number is the product of the other two. This implies a mathematical system of equations where each variable equates to the product of the other two, which can be written as:
- a = b × c
- b = a × c
- c = a × b
To find such triples, we can start by noting that if any of two variables are zero, the third must also be zero, giving us the trivial solution (0, 0, 0). If none of the variables are zero, we have a = b × c = b/b × c/c = 1/1 = 1, leading to the conclusion that a, b, and c must all be 1 or -1 in order to satisfy the condition since 1 and -1 are the only integers that are their own multiplicative inverses. Thus, we have four non-trivial solutions where each variable is 1 or -1: (1, 1, 1), (-1, -1, 1), (-1, 1, -1), and (1, -1, -1). Combining the trivial and non-trivial solutions, we get a total of 5 solutions.