Solution:
Notice that $2002+2$ divides $2002^{2}-2^{2}$, so any common divisor of $2002+2$ and $2002^{2}+2$ must divide $(2002^{2}+2)-(2002^{2}-2^{2})=6$. On the other hand, every number in the sequence is even, and the $n$th number is always congruent to $1^{n}+2 \equiv 0$ modulo $3$. Thus, $6$ divides every number in the sequence.