Let $f_{1}(x)=\sqrt{1-x}$, and for integers $n \geq 2$, let $f_{n}(x)=f_{n-1}(\sqrt{n^2 - x})$. If $N$ is the largest value of $n$ for which the domain of $f_{n}$ is nonempty, the domain of $f_{N}$ is $[c]$. What is $N+c$?
$\textbf{(A)}\ -226 \qquad \textbf{(B)}\ -144 \qquad \textbf{(C)}\ -20 \qquad \textbf{(D)}\ 20 \qquad \textbf{(E)}\ 144$
The domain of $f_{1}(x)=\sqrt{1-x}$ is defined when $x\leq1$. 
\[f_{2}(x)=f_{1}\left(\sqrt{4-x}\right)=\sqrt{1-\sqrt{4-x}}\]
Applying the domain of $f_{1}(x)$ and the fact that square roots must be positive, we get $0\leq\sqrt{4-x}\leq1$. Simplifying, the domain of $f_{2}(x)$ becomes $3\leq x\leq4$. 
Repeat this process for $f_{3}(x)=\sqrt{1-\sqrt{4-\sqrt{9-x}}}$ to get a domain of $-7\leq x\leq0$.  
For $f_{4}(x)$, since square roots must be nonnegative, we can see that the negative values of the previous domain will not work, so $\sqrt{16-x}=0$. Thus we now arrive at $16$ being the only number in the of domain of $f_4 x$ that defines $x$. However, since we are looking for the largest value for $n$ for which the domain of $f_{n}$ is nonempty, we must continue checking until we arrive at a domain that is empty. 
We continue with $f_{5}(x)$ to get a domain of $\sqrt{25-x}=16 \implies x=-231$. Since square roots cannot be negative, this is the last nonempty domain. We add to get $5-231=\boxed{\textbf{(A)}\ -226}$.