G2.1 If 191 is the difference of two consecutive perfect squares, find the value of the smallest square number, $a$.
Let $a=t^{2}$, the larger perfect square is $(t+1)^{2}$
$$
\begin{array}{l}
(t+1)^{2}-t^{2}=191 \\
2 t+1=191 \\
t=95 \\
a=95^{2}=9025
\end{array}
$$