Let $t$ be TNYWR.

The pair $(x, y)=(a, b)$ is a solution of the system of equations

$$
\begin{aligned}
& x^{2}+4 y=t^{2} \\
& x^{2}-y^{2}=4
\end{aligned}
$$

If $b>0$, what is the value of $b$ ?
Since $(x, y)=(a, b)$ satisfies the system of equations, then

$$
\begin{aligned}
& a^{2}+4 b=t^{2} \\
& a^{2}-b^{2}=4
\end{aligned}
$$

Subtracting the second equation from the first, we obtain successively

$$
\begin{aligned}
\left(a^{2}+4 b\right)-\left(a^{2}-b^{2}\right) & =t^{2}-4 \\
b^{2}+4 b & =t^{2}-4 \\
b^{2}+4 b+4 & =t^{2} \\
(b+2)^{2} & =t^{2} \\
b+2 & = \pm t \\
b & =-2 \pm t
\end{aligned}
$$

Since the answer to (b) is 16 , then $t=16$.

Therefore, $b=-2+t=14$ or $b=-2-t=-18$.

Since $b>0$, then $b=14$.

ANSWER: $6,16,14$