For how many integers $x$ is the expression $\frac{\sqrt{75-x}}{\sqrt{x-25}}$ equal to an integer?
For the square root in the numerator to be defined, we need $75-x \geq 0$ or $x \leq 75$, and for the square root in the denominator to be defined, we need $x-25 \geq 0$ or $25 \leq x$. Therefore, for the expression to be an integer, we must have $25 \leq x \leq 75$.

Suppose $\frac{\sqrt{75-x}}{\sqrt{x-25}}$ is equal to some integer $n$. Then $\sqrt{75-x}=n \sqrt{x-25}$.

Squaring both sides, we get $75-x=n^{2}(x-25)$ or $75-x=x n^{2}-25 n^{2}$. This equation can be rearranged to get $25 n^{2}+75=x\left(n^{2}+1\right)$ and then solved for $x$ to get $x=\frac{25 n^{2}+75}{n^{2}+1}$.

Rearranging, we have

$$
x=\frac{25 n^{2}+75}{n^{2}+1}=\frac{25 n^{2}+25+50}{n^{2}+1}=\frac{25\left(n^{2}+1\right)}{n^{2}+1}+\frac{50}{n^{2}+1}=25+\frac{50}{n^{2}+1}
$$

Observe that $n^{2}+1>0$ for all $n$, so the cancellation above is justified.

Since $x$ and 25 are both integers, $\frac{50}{n^{2}+1}$ must also be an integer, so 50 is a multiple of $n^{2}+1$. Since $n^{2}+1$ is a divisor of 50 and it must be positive, we have that $n^{2}+1$ must be one of 1,2 , $5,10,25$, or 50 .

Setting $n^{2}+1$ equal to each of these values and solving gives

$$
\begin{array}{ll}
n^{2}+1=1 & \Rightarrow n=0 \\
n^{2}+1=2 & \Rightarrow n= \pm 1 \\
n^{2}+1=5 & \Rightarrow n= \pm 2 \\
n^{2}+1=10 & \Rightarrow n= \pm 3 \\
n^{2}+1=25 & \Rightarrow n= \pm \sqrt{24} \\
n^{2}+1=50 & \Rightarrow n= \pm 7
\end{array}
$$

Note that $n$ is the ratio of two square roots, which means it must be nonnegative. As well, $\sqrt{24}$ is not an integer, so this means $n=0, n=1, n=2, n=3$, or $n=7$.

Substituting these values of $n$ into $x=\frac{25 n^{2}+75}{n^{2}+1}$, we get that the possible values of $x$ are 75 , $50,35,30$, and 26 , of which there are 5 .

ANSWER: 5