If the square root of \(p^2\) is an integer greater than 1, which of the following must be true?

Vincen wrote: ↑

Mon May 18, 2020 7:03 am

If the square root of \(p^2\) is an integer greater than 1, which of the following must be true?

I. \(p^2\) has an odd number of positive factors

II. \(p^2\) can be expressed as the product of an even number of positive prime factors

III. \(p\) has an even number of positive factors

A. I
B. II
C. III
D. I and II
E. II and III

[spoiler]OA=D[/spoiler]

Since p^2 > 1, p > 1 or p < -1. Let’s consider p > 1 first.

If p > 1, then p^2 always has an odd number of positive factors. So statement I is true. Statement II is also true since every distinct prime factor of p^2 must come in pairs or an even amount. However, statement III is not necessarily true since p itself can be a perfect square so p could have an odd number of positive factors.

Now, let’s consider p < -1.

If p < -1, statements I and II will still be true since p^2 = |p|^2 (notice that |p| > 1, so it behaves the same as p > 1). We don’t have to consider statement III because it’s proven false when p > 1 already.

Answer: D