3. The integer sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n+2}=a_{n+1}^{2}+a_{n}^{2}(n \geqslant 1) .
$$

If the positive integer $m$ satisfies $a_{m}=2005$, then the set of all possible $m$ is ( ).
(A) $\{1,2\}$
(B) $\{1,2,3\}$
(C) $\{1,2,3,4\}$
(D) $\{1,2,3,4,5\}$
3. B.

Let $a_{1}=2005$, then $m=1$ is possible;
Let $a_{2}=2005$, then $m=2$ is possible;
Let $a_{1}=18, a_{2}=41$. Then
$$
a_{3}=a_{1}^{2}+a_{2}^{2}=324+1681=2005,
$$

so $m=3$ is also possible.
If $m=4$ is possible, then we have
$$
2005=a_{2}^{2}+\left(a_{1}^{2}+a_{2}^{2}\right)^{2}\left(a_{1}, a_{2} \in \mathbf{N}_{+}\right) \text {. }
$$

Since $45^{2}=2025$, we know that $a_{1}^{2}+a_{2}^{2} \leqslant 44$, hence $a_{1}, a_{2} \leqslant 6$.
Thus, $\left(a_{1}^{2}+a_{2}^{2}\right)^{2}=2005-a_{2}^{2} \in[1969,2004]$.
However, $44^{2}=1936<1969<2004<45^{2}$, which is a contradiction.
When $m \geqslant 5$, similarly, it is not possible.