Solving $\divisibility(a,b)+\operatorname{lcm}(a,b)=9+b$

Reduction. Let $\delta=\divisibility(a,b)$, $a=\delta a'$, $b=\delta b'$, with $\divisibility(a',b')=1$. Then $\operatorname{lcm}(a,b)=\delta a'b'$, and the equation becomes
$$\delta+\delta a'b'=9+\delta b',$$
i.e.
$$\delta(1+a'b'-b')=9.$$
So $\delta\mid9$.

Case analysis. $\delta\in\{1,3,9\}$.

• $\delta=1$: $1+a'b'-b'=9$, so $a'b'-b'=8$, i.e. $b'(a'-1)=8$. With $\divisibility(a',b')=1$, solve the divisor cases.
• $\delta=3$: $3(1+a'b'-b')=9$, so $a'b'-b'=2$, i.e. $b'(a'-1)=2$.
• $\delta=9$: $1+a'b'-b'=1$, so $a'b'=b'$, i.e. $b'(a'-1)=0$, hence $a'=1$ (since $b'>0$). Then $a=\delta=9$ and $b=9b'$ arbitrary.

Conclusion. The full solution set is obtained by checking the finitely many divisor cases for each $\delta$. For $\delta=9$, the solutions are $a=9$, $b\equiv0\pmod9$.