Problem (Algebra):
Let $T=5$. Compute the smallest positive integer $n$ such that there are at least $T$ positive integers in the domain of $f(x)=\sqrt{-x^{2}-2 x+n}$.

Solution:
Completing the square under the radical yields $\sqrt{n+1-(x+1)^{2}}$. The larger zero of the radicand is $-1+\sqrt{n+1}$, and the smaller zero is negative because $-1-\sqrt{n+1}<0$, so the $T$ positive integers in the domain of $f$ must be $1,2,3, \ldots, T$. Therefore $-1+\sqrt{n+1} \geq T$. Hence $\sqrt{n+1} \geq T+1$, and $n+1 \geq(T+1)^{2}$. Therefore $n \geq T^{2}+2 T$, and substituting $T=5$ yields $n \geq 35$. So $n=\mathbf{3 5}$ is the smallest such value.

Final answer: 35