Proizvolov V.V.

Find all natural numbers $a$ and $b$ such that $\left(a+b^{2}\right)\left(b+a^{2}\right)$ is an integer power of two.
Each bracket is a power of two. Therefore, the numbers $a$ and $b$ have the same parity. We can assume that $a \geq b$. Let's consider two cases.

1) $a=b$. Then $a^{2}+a=a(a+1)$ is a power of two. The numbers $a$ and $a+1$ are powers of two with different parities, hence $a=1$.
2) $a>b$. Then $a^{2}+b=2^{m}>b^{2}+a=2^{n}$. We have $\left(2^{m-n}-1\right) \cdot 2^{n}=a^{2}-b^{2}+b-a=(a-b)(a+b-1)$. $a-b$ is even, and $a+b-1$ is odd, therefore,

$a-b=2^{n}=a+b^{2}$. This is a contradiction.

## Answer

$a=b=1$.