Given Information
1. $q$ is an integer.
2. $q$ is the square root of $p$. This means: $\(q=\sqrt{p}\)$.

From $\(q=\sqrt{p}\)$, we can derive the relationship between $p$ and $q$ by squaring both sides:

$$
\(p=q^2\)
$$


Since $q$ is an integer, $p$ must be the square of an integer, meaning $p$ must be a perfect square integer. Also, since $p$ is a square, $p$ must be non-negative.

We need to determine which of the options must be an integer based on this information.

Analysis of Options
(A) $\(\frac{p}{q}\)$

Substitute $\(p=q^2\)$ into the expression:

$$
\(\frac{p}{q}=\frac{q^2}{q}\)
$$


Since $\(q=\sqrt{p}, p\)$ must be non-negative.
- Case 1: $\(q \neq 0\)$ (and thus $\(p \neq 0\)$ )

$$
\(\frac{q^2}{q}=q\)
$$


Since $q$ is given to be an integer, $\(\frac{p}{q}\)$ is an integer.
- Case 2: $q=0$ If $q=0$, then $\(p=0^2=0\)$. The expression becomes $\(\frac{0}{0}\)$, which is undefined.

- Conclusion: If we assume the question implies $q \neq 0$ for the division to be defined, then (A) must be an integer (it is equal to $q$ ). However, if $q$ can be 0 , it is not an integer. In standard quantitative testing, we usually assume division by zero is avoided unless explicitly allowed. Given the options, and assuming the question intends for the expression to be defined, (A) is the most likely intended answer, as $p / q=q$.

Assuming $\(q \neq 0\)$ for a defined quotient: (A) must be an integer.
(B) $\(\frac{q}{p}\)$

Substitute $\(p=q^2\)$ into the expression:

$$
\(\frac{q}{p}=\frac{q}{q^2}=\frac{1}{q}\)
$$


For $\(\frac{1}{q}\)$ to be an integer, $q$ must be 1 or -1 .

- If $q=2$ (an integer), then $\(\frac{q}{p}=\frac{1}{2}\)$ (not an integer).
- If $q=-2$ (an integer), then $\(\frac{q}{p}=-\frac{1}{2}\)$ (not an integer).

Since $\(\frac{q}{p}\)$ is not an integer for all possible integer values of $q$ (e.g., $q=2$ ), it does not must be an integer.

(C) $\(\sqrt{q}\)$
$q$ is an integer. $\sqrt{q}$ is only an integer if $q$ is a perfect square (e.g., $\(q=4, \sqrt{q}=2\)$ ).
- If $q=3$ (an integer), then $\(\sqrt{q}=\sqrt{3}\)$ (not an integer).

It does not must be an integer.
(D) $\(\sqrt{p}\)$

From the given information, we have $\(q=\sqrt{p}\)$. Since $q$ is given to be an integer, $\(\sqrt{p}\)$ must be an

Final Selection
Both (A) and (D) are derived from the same premise $\(q=\sqrt{p}\)$ and $\(p=q^2\)$.

- (D) $\(\sqrt{p}=q\)$, which is given as an integer. (D must be an integer.)
- (A) $\(\frac{p}{q}=\frac{q^2}{q}=q\)$. Since $q$ is an integer, $\(\frac{p}{q}\)$ is an integer (assuming $\(q \neq 0\)$ ). (A must be an integer, provided the quotient is defined.)

Since the question asks what must be an integer and the definitions align perfectly, we select both (A) and (D), assuming the context prevents division by zero.

The correct choices are (A) and (D).