3-3. Find such non-zero, unequal integers $a, b, c$ that the expression

$$
x(x-a)(x-b)(x-c)+1
$$

can be factored into a product of two polynomials with integer coefficients.
Solution 3. Let $x(x-a)(x-b)(x-c)+1=P(x) Q(x)$, where $P(x)$ and $Q(x)$ are polynomials with integer coefficients. It is clear that $P(x)$ 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 $x(x-a)(x-b)(x-c)+1:\left(x^{2}-3 x+1\right)^{2},\left(x^{2}+3 x+1\right)^{2}$, $\left(x^{2}+x-1\right)^{2},\left(x^{2}-x-1\right)^{2}$.

Solution 4. Answer: $(0,0),(0,1),(1,0),(1,2),(2,1),(2,2)$.

Consider the given equation as a quadratic equation in $x$:

$$
x^{2}-(y+1) x+y^{2}-y=0
$$

The discriminant of this equation is $-3 y^{2}+6 y+1$. It is negative for $y \geqslant 3$ and for $y \leqslant-1$. Therefore, for $y$ we get three possible values: $0,1,2$. For each of these values, we get an equation that is easily solved.