HMMT February

Notice that $2002+2$ divides $2002^{2}-2^{2}$, so any common divisor of $2002+2$ and $2002^{2}+2$ must divide $\left(2002^{2}+2\right)-\left(2002^{2}-2^{2}\right)=6$. 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.