Product of Difference of 3 and Difference of 2?

Let $n$ be the number that can be written as $a^2-1$ (product of $a-1$ and $a+1$ with a difference of $2$ ) and $b^2-3b$ (product of two numbers $b$ and $b-3$ with a difference of $3$ ). Then $b^2-3b+1-a^2=0$ , or $b=\dfrac{3+\sqrt {(2a)^2+5}}{2}$ . Now the only two perfect squares with a difference of $5$ are $4$ and $9$ . So $a=0$ and hence $n=0$ . Since this is the only number, the required sum is also $\boxed 0$