4. Let $a$ and $b$ be two coprime natural numbers. Then the greatest common divisor of $a^{2}+$ $b^{2}$ and $a^{3}+b^{3}$ is ( ).
(A) 1
(B) 2
(C) 1 or 2
(D) Possibly greater than 2
4.C.

If we take $a=2, b=3$, then $a^{2}+b^{2}=13, a^{3}+b^{3}=35$. 13 and 35 are coprime.

If we take $a=3, b=5$, then $a^{2}+b^{2}$ and $a^{3}+b^{3}$ are both even, they have a common factor of 2.
Now assume $a^{2}+b^{2}$ and $a^{3}+b^{3}$ have a common factor of 4, then because
$$
a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}\right)-a b(a+b)
$$

and $a, b$ are coprime, it must be that $2 \mid (a+b)$. Thus, $4 \mid (a+b)^{2}$.
And $(a+b)^{2}=a^{2}+b^{2}+2ab$, from which we get $4 \mid 2ab$, i.e., $2 \mid ab$.
Therefore, $2 \mid a$ or $2 \mid b$.
In either case, it leads to $(a, b) \geqslant 2$, which contradicts the fact that $a$ and $b$ are coprime. Similarly, we can conclude that $a^{2}+b^{2}$ and $a^{3}+b^{3}$ do not have any prime common factor greater than 2.