3-4. Find the integer $a$, for which

$$
(x-a)(x-10)+1
$$

can be factored into the product $(x+b)(x+c)$ of two factors with integer $b$ and $c$.
Solution 4. Let $(x-a)(x-10)+1=(x+b)(x+c)$. By setting $x=-b$, we get $(b+a)(b+10)=-1$. Since $a$ and $b$ are integers, $b+a$ and $b+10$ are also integers. The number -1 can be represented as the product of two factors in two ways. Accordingly, we get two cases: 1) $b+10=1$ and $b+a=-1$; 2) $b+10=-1$ and $b+a=1$. Therefore, $a=8$ or $a=12$. In the first case, $(x-8)(x-10)+1=(x-9)^{2}$, and in the second case, $(x-12)(x-10)+1=(x-11)^{2}$.