x and y are positive integers. If x is a prime factor of y^2

If \(x\) and \(y\) are positive integers and \(x\) is a prime factor of \(y^2\), then which of the following MUST be true?

A. \(x\) is even
B. \(x = y^2\)
C. \(\frac{y}{x}\) is even
D. \(\frac{x}{y} \le 1\)
E. \(x+1\) is also a prime factor of \(y^2\)


Given that \(x\) is a prime factor of \(y^2\), and \(y\) is a positive integer, then \(x\) must be a factor of \(y\) as well - how else could \(x\) appear in the prime factorization of \(y^2\) (which is \(y*y\)) if it's not a factor of \(y\)? Accordingly, as \(x\) is a factor of \(y\) (and \(y\) is a positive integer), it follows that \(x \le y\). Therefore, \(\frac{x}{y} \le 1\), so option D must be true.

Alternative approach:

It's important to note that the question asks for what "MUST be true", not "COULD be true". In such questions, if we can demonstrate that a statement is NOT true for a specific set of numbers, it means that this statement is not always true, and therefore, not the correct answer.

A. \(x\) is even. This is not necessarily true. For example, if \(x=3\) (which is odd) and \(y=3\), then this statement doesn't hold.

B. \(x=y^2\). This is never true. As \(x\) is prime, it cannot be equal to the square of any integer.

C. \(\frac{y}{x}\) is even. This is not necessarily true. For instance, if \(x=3\) and \(y=3\), then \(\frac{y}{x}=1\), which is odd.

D. \(\frac{x}{y} \le 1\). This is always true, as explained above or via process of elimination.

E. \(x+1\) is also a prime factor of \(y^2\). This is not necessarily true. For example, if \(x=2\) and \(y=2\), then \(x+1=3\) and 3 is not a factor of \(y^2=4\).


Answer: D