1. Given that $a$, $b$, and $c$ are all integers, and it always holds that
$$
(x-a)(x-10)+1=(x+b)(x+c) \text {. }
$$

then the integer $a=$ $\qquad$
II. 1.8 or 12.

From the given, we have
$$
(x-a)(x-10)+1=(x+b)(x+c) \text {. }
$$

Then $x^{2}-(a+10) x+10 a+1$
$$
=x^{2}+(b+c) x+b c \text {. }
$$

Thus, $\left\{\begin{array}{l}-(a+10)=b+c, \\ 10 a+1=b c .\end{array}\right.$
Therefore, $-99=10 b+10 c+b c$, which means
$$
c=\frac{-10 b-100+1}{b+10}=-10+\frac{1}{b+10} \text {. }
$$

Since $a, b, c$ are integers, we have $b+10= \pm 1$, i.e., $b=-9$ or -11.
Thus, $c=-9$ or -11.
Hence, when $b=c=-9$, $a=8$;
When $b=c=-11$, $a=12$.