308. a) Find an integer \( a \) such that

\[
(x-a)(x-10)+1
\]

can be factored into the product \((x+b)(x+c)\) of two factors with integer \( b \) and \( c \).

b) Find such non-zero, distinct integers \( a, b, c \) that the polynomial of the fourth degree with integer coefficients

\[
x(x-a)(x-b)(x-c)+1
\]

can be represented as the product of two other polynomials with integer coefficients.
308. a) Let $(x-a)(x-10)+1=(x+b)(x+c)$. By substituting $x=-b$ into both sides of the equation, we get:

$$
(-b-a)(-b-10)+1=(-b+b)(-b+c)=0
$$

From this,

$$
(b+a)(b+10)=-1
$$

Since $a$ and $b$ are integers, $b+a$ and $b+10$ are also integers. However, -1 can be represented as the product of two integers in only one way: $-1=1 \cdot(-1)$. Therefore, there are only two possibilities:

1) $b+10=1, b=-9$; then $b+a=-9+a=-1$, i.e., $a=8$:

$$
(x-8)(x-10)+1=(x-9)^{2}
$$

2) $b+10=-1, b=-11$; then $b+a=-11+a=1$, i.e., $a=12$:

$$
(x-12)(x-10)+1=(x-11)^{2}
$$

b) Since a polynomial of the fourth degree can be factored either as the product of a polynomial of the first degree and a polynomial of the third degree or as the product of two polynomials of the second degree, we need to consider two cases separately:

$$
\text { A) } x(x-a)(x-b)(x-c)+1=(x+p)\left(x^{3}+q x^{2}+r x+s\right)
$$

(The coefficients of $x$ in the first factor and of $x^{3}$ in the second factor on the right-hand side are both 1 or both -1, since the product of these factors must have a coefficient of 1 for $x^{4}$, which is the same as the coefficient of $x^{4}$ in the expression $x(x-a)(x-b) \times$ $\times(x-c)+1$; the equality $x(x-a)(x-b)(x-c)+1=$ $=\left(-x+p_{1}\right)\left(-x^{3}+q_{1} x^{2}+r_{1} x+s_{1}\right)$ can be transformed into the form (*) by multiplying both factors on the right-hand side by -1).

By substituting $x=0, x=a, x=b$, and $x=c$ into the equation (*), and considering that 1 can be factored in only two ways $1=1 \cdot 1=(-1) \cdot(-1)$, we get that the four distinct numbers $0+p=p, a+p, b+p$, and $c+p$ (we remind that the numbers $0, a, b, c$ are all distinct) can only take two values +1 and -1, which is impossible.

B) $x(x-a)(x-b)(x-c)+1=\left(x^{2}+p x+q\right)\left(x^{2}+r x+s\right)$.

From this, as above, we get that for $x=0, x=a, x=b$, and $x=c$, both polynomials, $x^{2}+p x+q$ and $x^{2}+r x+s$, take the values 1 or -1. However, a quadratic polynomial $x^{2}+p x+q$ cannot take the same value $\alpha$ at three different values of $x$ (otherwise, the quadratic equation $x^{2}+p x+q-\alpha=$ 0 would have three different roots), which implies that at two of the four values $x=0, x=a, x=b, x=c$, this polynomial takes the value 1, and at the other two values, it takes the value -1. Suppose that $0^{2}+p \cdot 0+q=q=1$, and let $x=a$ be the value of $x=a$, $x=b$, and $x=c$ at which this polynomial takes the same value 1; in this case, at $x=b$ and $x=c$, it takes the value -1. Thus, we have:

$$
a^{2}+p a+1=1, b^{2}+p b+1=-1, c^{2}+p c+1=-1
$$

From $a^{2}+p a=a(a+p)=0$ it follows that $a+p=0, p=-a$ (since, by assumption, $a \neq 0$). Thus, the last two equations become

$$
b^{2}-a b=b(b-a)=-2, \quad c^{2}-a c=c(c-a)=-2
$$

Subtracting the second equation from the first, we get:

$$
\begin{aligned}
& b^{2}-a b-c^{2}+a c=(b-c)(b+c)-a(b-c)= \\
&=(b-c)(b+c-a)=0
\end{aligned}
$$

from which, since $b \neq c$, we have: $b+c-a=0, a=b+c, b-a=$ $=-c, c-a=-b$. Now from the equation

$$
b(b-a)=-b c=-2
$$

we get the following values for $b, c$, and $a$:

$$
\begin{gathered}
b=1, \quad c=2, \quad a=b+c=3 \\
x(x-a)(x-b)(x-c)+1=x(x-3)(x-1)(x-2)+1= \\
b=-1, c=-2, a=b+c=-3 \\
x(x-a)(x-b)(x-c)+1=x(x+3)(x+1)(x+2)+1= \\
\quad, \quad\left(x^{2}-3 x+1\right)^{2} \\
\\
\quad=\left(x^{2}+3 x+1\right)^{2}
\end{gathered}
$$

Similarly, if $x^{2}+p x+q$ takes the value -1 at $x=0$ and $x=a$, and the value +1 at $x=b$ and $x=c$, then we have: $q=-1, \quad a^{2}+p a-1=-1, b^{2}+p b-1=1, c^{2}+p c-1=1$, from which

$$
\begin{gathered}
p=-a, b(b-a)=c(c-a)=2, b^{2}-a b-c^{2}+a c=0 \\
(b-c)(b+c-a)=0, \quad a=b+c, \quad b-a=-c, \quad-b c=2
\end{gathered}
$$

Thus, we get two more possible systems of values for $a, b$, and $c$:

$$
\begin{gathered}
b=2, \quad c=-1, \quad a=b+c=1 \\
x(x-a)(x-b)(x-c)=x(x-1)(x-2)(x+1)+1= \\
=\left(x^{2}-x-1\right)^{2} \\
\cdot b=1, \quad c=-2, \quad a=b+c=-1 \\
x(x-a)(x-b)(x-c)=x(x+1)(x-1)(x+2)+1= \\
=\left(x^{2}+x-1\right)^{2}
\end{gathered}
$$

Note. Another solution to this problem is provided in the final part of the solution to problem 309 b).