16. (25 points) Given that $a$, $b$, and $c$ are integers, satisfying
$$
c>0, a+b=3, c^{2}-2 c-a b=-2 \text {. }
$$

If the equation in $x$
$$
d x^{2}+(x+d) x+a b+d=0
$$

has only one solution, find the value of $d$.
16. It is known that $a$ and $b$ are the roots of the equation
$$
x^{2}-3 x+c^{2}-2 c+2=0
$$

with $\Delta=9-4\left(c^{2}-2 c+2\right) \geqslant 0$, then $4 c^{2}-8 c-1 \leqslant 0$.
Given that $c$ is a positive integer, solving yields $c=1,2$.
When $c=1$, $a$ and $b$ have no integer solutions;
When $c=2$, $a=1, b=2$ or $a=2, b=1$.
Thus, the original equation becomes
$$
d x^{2}+(2+d) x+2+d=0 \text {. }
$$

When $d=0$, $x=-1$;
When $d \neq 0$, equation (1) has equal roots, and
$$
\Delta=(2+d)^{2}-4 d(2+d)=0 \text {. }
$$

Solving yields $d=\frac{2}{3},-2$.
In summary, the values of $d$ are $0, \frac{2}{3},-2$.