smc

Change the equation to base 10: $$A^2+3A+2+4B+3=6A+6B+9$$ $$A^2-3A-2B-4=0$$ Either $B=A+1$ or $B=A-1$, so either $A^2-5A-6,B=A+1$ or $A^2-5A-2,B=A-1$. The second case has no integer roots, and the first can be re-expressed as $(A-6)(A+1)=0,B=A+1$. Since A must be positive, $A=6,B=7$ and $A+B=\boxed{13}$