It is obvious $a $ can't be negative number .Let $a$ is positive number . Define positive sequence $(x_n)$ such that $a.2013^n+b=(x_n)^2$ It is obvious that $(x_n)$ is increasing. $(2013.x_n)^2-x_{n+2}^2=b.(2013^2-1)$ Let $S=b.(2013^2-1)$ $S$ could be written difference of two square infinitly ways.So $ S$ is 0 (becase of the finite factorization.) So $b=0$ If $a.2013^n$ square then $a.2013^{n+1}$ isn't square .Contradiction .Thus $a=0$