G10.1 It is given that $3 \times 4 \times 5 \times 6=19^{2}-1$
$$
\begin{array}{l}
4 \times 5 \times 6 \times 7=29^{2}-1 \\
5 \times 6 \times 7 \times 8=41^{2}-1 \\
6 \times 7 \times 8 \times 9=55^{2}-1
\end{array}
$$

If $a^{2}=1000 \times 1001 \times 1002 \times 1003+1$, find the value of $a$.
$\begin{array}{l}19=4 \times 5-1 ; 29=5 \times 6-1 ; 41=6 \times 7-1 ; 55=5 \times 6-1 \\ a^{2}=(1001 \times 1002-1)^{2}-1+1=(1001 \times 1002-1)^{2} \\ a=1003002-1=1003001\end{array}$