6. For the equation about integers $a, b$: $\sqrt{a-1}+\sqrt{b-1}=\sqrt{a b+k}(k \in \mathbb{Z})$, if there is only one ordered real solution, then $k=$ $\qquad$
Solution: 0.
From the given equation, squaring both sides and rearranging yields $(a-1)(b-1)-2 \sqrt{(a-1)(b-1)}+1+k \neq 0$
$$
\text { we get }(\sqrt{(a-1)(b-1)}-1)^{2}=-k
$$

When $k \geq 1$, there is no solution.
When $k=0$, there is only one solution $(a, b)=(2,2)$.
When $k \leq-1$, $\sqrt{(a-1)(b-1)}=1 \pm \sqrt{-k}$
Squaring both sides gives $(a-1)(b-1)=1-k \pm 2 \sqrt{-k}$.
Thus, $\sqrt{-k}$ is an integer. Therefore, $(a-1)(b-1)=(1 \pm \sqrt{-k})^{2}$ has more than one solution.
In summary, $k=0$.