6.4.1 * Find all integer triples $(a, b, c)$ such that $0<|a|<|b|<|c|$, and the polynomial
$$
f(x)=x(x-a)(x-b)(x-c)+1
$$

is reducible in $\mathbf{Z}[x]$ (i.e., can be factored into the product of two polynomials with lower degrees and integer coefficients).
Let $f(x)=g(x) h(x), g, h \in \mathbf{Z}[x]$ be monic polynomials, and $1 \leqslant \operatorname{deg} g \leqslant \operatorname{deg} h$. We discuss the following two cases.
(1) If $\operatorname{deg} g=1$, then there exists $d \in \mathbf{Z}$, such that $d(d-a)(d-b)(d-c)=-1$, but -1 cannot be expressed as the product of 4 integers, which is a contradiction.
(2) If $\operatorname{deg} g=\operatorname{deg} h=2$, let
$$
f(x)=\left(x^{2}+p x+q\right)\left(x^{2}+r x+t\right) .
$$

By setting $x=0$ and $x=c$, we know $q t=\left(c^{2}+p c+q\right)\left(c^{2}+r c+t\right)=1$, thus $q=t= \pm 1, c^{2}+$ $p c+q=c^{2}+r c+t= \pm 1$. At this point, since $|c| \geqslant 3$, we have $\left|c^{2}+p c\right|=|c+p| \cdot|c|$. Therefore, if $c+p \neq 0$, then $\left|c^{2}+p c+q\right| \geqslant\left|c^{2}+p c\right|-|q| \geqslant 3-1=2$, which is a contradiction. Hence, $c=-p$. Similarly, $c=-r$. Therefore, we should have
$$
f(x)=x(x-a)(x-b)(x-c)+1=\left(x^{2}-c x \pm 1\right)^{2} .
$$

By comparing coefficients, we know $a+b+c=2 c, a b+b c+c a=c^{2} \pm 2$, thus $c=a+b, a b+(a+b)^{2}=(a+b)^{2} \pm 2$, i.e., $a b= \pm 2$, hence $a= \pm 1, b=+2$. Combining with $c=a+b$, we know $(a, b, c)=(1,2,3)(-1,-2,-3),(-1,2,1)$ or $(1,-2,-1)$.