2. Find all pairs of natural numbers $a$ and $b$ such that

$$
\operatorname{HCF}(a+1, b+1)=a^{2}-b^{2}
$$
2. Let $a+1=d x$ and $b+1=d y$, where $d, x, y$ are natural numbers, $x>y$ and GCD $(x, y)=1$. Then LCM $(a+1, b+1)=d x y=(d x-1)^{2}-(d y-1)^{2}=$ $d(x-y)(d x+d y-2)$ and hence $d x+d y-2=\frac{x y}{x-y}$. Since $x-y$ is coprime with both $x$ and $y$, it must be $x-y=1$. The previous equation becomes $d(2 x-1)-2=x(x-1)$, so $2 x-1\left|x^{2}-x+2\right|(2 x-1)^{2}+7$, thus $2 x-1 \mid 7$, i.e., $x=1$ or $x=4$. The first possibility obviously fails, and the second gives $d=2$ and $(a, b)=(7,5)$.