10.33. a) Find the integer $a$ for which $(x-a) \times$ $\times(x-10)+1$ can be factored into the product $(x+b)(x+c)$ of two factors with integer $b$ and $c$.

b) Find such non-zero and distinct integers $a, b, c$ that the expression $x(x-a)(x-b) \times$ $\times(x-c)+1$ can be factored into the product of two polynomials with integer coefficients.

## 10.9. Interpolation Polynomials
10.33. a) 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 integers in two ways. Accordingly, we have 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, $(x-12)(x-10)+1=(x-11)^{2}$.

b) Let $x(x-a)(x-b)(x-c)+1=P(x) Q(x)$, where $P(x)$ and $Q(x)$ are polynomials with integer coefficients. Clearly, $P(x)$ and $Q(x)$ are polynomials with leading coefficient 1. For $x=0, x=a$, $x=b$, and $x=c$, we have $P(x) Q(x)=1$, i.e., either $P(x)=1$ and $Q(x)=1$, or $P(x)=-1$ and $Q(x)=-1$. In both cases, $P(x)-Q(x)=0$. The degree of the polynomial $P(x)-Q(x)$ is strictly less than four, so $P(x)=Q(x)$ for all $x$. Thus, $x(x-a)(x-b)(x-c)=P^{2}(x)-1=(P(x)+1)(P(x)-1)$. Therefore, $P(x) \pm 1=x(x-a)$ and $P(x) \mp 1=(x-b)(x-c)$, i.e., $x(x-a)-(x-b)(x-c)= \pm 2$ (we do not distinguish solutions that differ only by the permutation of the numbers $a, b, c$). Consequently, $a=b+c$ and $bc=\mp 2$. As a result, we obtain the following sets of values $(a, b, c):(1,2,3),(-1,-2,-3),(1,-2,-1),(2,-1,1)$. These correspond to the following factorizations of the polynomial $x(x-a)(x-b) \times$ $\times(x-c)+1:\left(x^{2}-3 x+1\right)^{2},\left(x^{2}+3 x+1\right)^{2},\left(x^{2}+x-1\right)^{2},\left(x^{2}-x-1\right)^{2}$.