For how many integer values of $x$ is $\sqrt{25-x^{2}}$ equal to an integer?
(A) 7
(B) 6
(C) 5
(D) 3
(E) 2

## Part C: Each correct answer is worth 8.
For $\sqrt{25-x^{2}}$ to be defined, $25-x^{2} \geq 0$, or $x^{2} \leq 25$, or $-5 \leq x \leq 5$. So we make a table of $x$ and $25-x^{2}$ and check to see when $25-x^{2}$ is a perfect square, ie. $\sqrt{25-x^{2}}$ is an integer.

| $x$ | $25-x^{2}$ | Perfect Square? |
| :---: | :---: | :---: |
| 0 | 25 | Yes |
| $\pm 1$ | 24 | No |
| $\pm 2$ | 21 | No |
| $\pm 3$ | 16 | Yes |
| $\pm 4$ | 9 | Yes |
| $\pm 5$ | 0 | Yes |

Therefore, there are 7 integer values for $x$ (namely $0, \pm 3, \pm 4, \pm 5$ ) that make $\sqrt{25-x^{2}}$ an integer.