1B. Determine all pairs of natural numbers $(a, b)$ such that $a \leq b$ and

$$
\left(a+\frac{6}{b}\right)\left(b+\frac{6}{a}\right)=25
$$
Solution. The given equation is successively equivalent to the equations

\[
\begin{aligned}
& a b+\frac{36}{a b}+12=25 \\
& (a b)^{2}-13 a b+36=0 \\
& (a b-4)(a b-9)=0
\end{aligned}
\]

According to this, $a b=4$ or $a b=9$. But, $a, b \in \mathbb{N}$ and $a \leq b$, so:

- from $a b=4$ it follows that $(a, b)=(1,4)$ or $(a, b)=(2,2)$,
- from $a b=9$ it follows that $(a, b)=(1,9)$ or $(a, b)=(3,3)$.