[ Factorization ] [ Methods for solving problems with parameters ]

Find such non-zero and unequal integers $a, b, c$ so that the expression $x(x-a)(x-b)(x-c) + 1$ can be factored into the product of two polynomials (of non-zero degree) with integer coefficients.
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 $P(x) Q(x)=1$, meaning either $P(x)=Q(x)=1$ or $P(x)=Q(x)=-1$. In both cases, $P(x)-Q(x)=0$. The degree of the polynomial $P(x)-Q(x)$ is less than four, and it equals zero at four points, so $P(x)=Q(x)$ for all $x$. Therefore, $x(x-a)(x-b)(x-c)=P^{2}(x)-1=(P(x)+1)(P(x)-1)$. Hence, $P(x) \pm 1=x(x-a)$ and $P(x) \mp 1=(x-b)(x-c)$, which means $x(x-a)-(x-b)(x-c)= \pm 2$ (we do not distinguish solutions that differ only by permutations of the numbers $a, b, c$). Consequently, $a=b+c$ and $-b c= \pm 2$. As a result, we obtain the following sets of values $(a, b, c):(3,2,1),(-3,-2,-1),(-1,-2,1),(1,2,-1)$. These correspond to the following factorizations of the polynomial $x(x-a)(x-b)(x-c)+1$: $\left(x^{2}-3 x+1\right)^{2},\left(x^{2}+3 x+1\right)^{2},\left(x^{2}+x-1\right)^{2},\left(x^{2}-x-1\right)^{2}$.