3. Let $n$ be a given natural number, find all positive pairs $a, b$ such that $x^{2}+a x+b$ is a factor of $a x^{2 n}+(a x+b)^{2 n}$.
3. Let $x=x_{0}$ be a root of $x^{2}+a x+b=0$.

Since $b>0$, hence $x_{0} \neq 0$. $x_{0}$ is also a root of $a x^{2 n}+(a x+b)^{2 n}$, so
$$
a x^{2 n}+\left(a x_{0}+b\right)^{2 n}=a x_{0}^{2}+\left(-x_{0}^{2}\right)^{2 n}=a x_{0}^{2 n}+x_{0}^{4 n}=0 .
$$

Thus, $a+x_{0}^{2 n}=0$, so, $x_{0}^{2 n}=-a=a(\cos \pi+\mathrm{isin} \pi) . x_{0}=\sqrt[2 n]{a}\left(\cos \frac{(2 k+1) \pi}{2 n}+\mathrm{i} \sin \frac{(2 k+1) \pi}{2 n}\right)$.
On the other hand, by the quadratic formula, $x_{0}=\frac{-a \pm \sqrt{4 b-a^{2}} \mathrm{i}}{2}$.
Therefore, $\sqrt[2 n]{a} \cos \frac{(2 k+1) \pi}{2 n}=-\frac{a}{2}$.

This implies $\cos \frac{(2 k+1) \pi}{2 n}<0, \frac{\pi}{2}+2 k \pi<\frac{(2 k+1) \pi}{2 n}<\frac{3 \pi}{2}+2 k \pi$,
thus $n<2 k+1<3 n$. Therefore, when $n=1$, there is no solution.
When $n \geqslant 2$, $a=\left[-2 \cos \frac{(2 k+1) \pi}{2 n}\right]^{\frac{k n}{2 n-1}}=\left[2 \cos \frac{(2 k+1) \pi}{2}\right]^{\frac{2 .}{2 n-1}}$,
$$
b=x_{0} \overline{x_{0}}=\left|x_{0}\right|^{2}=\sqrt[n]{a}=\left[2 \cos \frac{(2 k+1) \pi}{2 n}\right]^{\frac{2}{2 n-1}} .
$$

Therefore, when $n \geqslant 2, n \in \mathbf{N}$, the positive pairs $(a, b)$ are
$$
a=\left[2 \cos \frac{(2 k+1) \pi}{2 n}\right]^{\frac{z_{n}}{n-1}}, b=\left[2 \cos \frac{(2 k+1) \pi}{2 n}\right]^{\frac{2}{k-1}},
$$

where $n<2 k+1<3 n, k \in \mathbf{N}$, and when $n=1$, there is no solution.