[ Factorization $]$

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 $c$.

#
Let $(x-a)(x-10)+1=(x+b)(x+c)$.

First method. Setting $x=-b$, we get $(b+a)(b+10)=-1$. Therefore, either $b+10=1, b+a=-1$, which means $a=8$; or $b+10=-1, b+a=1$, which means $a=12$. In the first case, $(x-8)(x-10)+1=(x-9)^{2}$, and in the second case, $(x-12)(x-10)+1=(x-11)^{2}$.

Second method. Substituting $x=10$, we get $(b+10)(c+10)=1$. Therefore, $b+10=c+10= \pm 1$, which means either $b=c=-9$, or $b=c=-11$. Substituting $x=0$, we get $10 a+1=b c$, from which we find $a$.

## Answer

$a=8$ or 12.