Deconstructing the QuestionWe are told that for the positive integer \(x\), the product of any two distinct positive factors of \(x\) is odd.
Key idea: an odd product requires both factors to be odd, so \(x\) cannot have any even factor.
Step-by-stepIf \(x\) were even, then \(2\) would be a positive factor of \(x\).
Also \(1\) is a distinct positive factor.
Then \(2\cdot 1=2\) would be even, contradicting the condition.
So \(x\) must be odd.
Check the statements:
I. \(x\) is an odd prime.
Counterexample: \(x=9\) works (all factors are odd), but \(9\) is not prime. So I is not must.
II. \(x^2\) is odd.
Since \(x\) must be odd, \(x^2\) must be odd. So II must be true.
III. \(x\) is the square of an odd number.
Counterexample: \(x=15\) works (all factors are odd), but \(15\) is not a square. So III is not must.
Answer: II only