Problem 3. (6 points) Find all natural numbers $a$ and $b$, for which $\left(a+b^{2}\right)\left(b+a^{2}\right) \mathrm{e}$ is of the form $2^{n}$ for some natural number $n$.

(Note: The symbol $\mathrm{e}$ in the original problem seems to be a typographical error and should be ignored or considered as a multiplication symbol. The correct interpretation is $\left(a+b^{2}\right)\left(b+a^{2}\right) = 2^n$.)
Solution. First, we note that $a+b^{2}$ and $b+a^{2}$ are powers of two, and the numbers $a$ and $b$ have the same parity. We will consider two cases:

1) $a=b$. Then $a^{2}+a=a(a+1)$ is a power of two. The numbers $a$ and $a+1$ have different parity and are powers of two, so $a=1$.
2) $a>b$. Then $a^{2}+b=2^{m}>b^{2}+a=2^{n}$. We have

$$
\left(2^{m-n}-1\right) \cdot 2^{n}=a^{2}-b^{2}+b-a=(a-b)(a+b-1)
$$

The number $a-b$ is even, and the number $a+b-1$ is odd, so $a-b=2^{n}=a+b^{2}$. Contradiction.

Thus, the solution to the problem is $a=b=1$.