5-36 Try to find 3 distinct non-zero integers $a, b, c$, such that the algebraic expression $x(x-a)(x-b)(x-c)+1$

can be expressed as the product of two polynomials with integer coefficients.
[Solution] (1) If $d, e, f, g$ are integers, and
$$
x(x-a)(x-b)(x-c)+1=(x+d)\left(x^{3}+e x^{2}+f x+g\right)
$$

then let $x=0$, we get $d g=1, d= \pm 1$. By setting $x=a, b, c$ in sequence, we obtain
$$
\left\{\begin{array}{l}
a+d= \pm 1, \\
b+d= \pm 1, \\
c+d= \pm 1 .
\end{array}\right.
$$

If $d=1$, then the non-zero integers $a=b=c=-2$, which contradicts the given condition that $a, b, c$ are distinct; if $d=-1$, then $a=b=c=2$, which is also a contradiction. Therefore, equation (1) cannot hold.
(2) If $d, e, f, g$ are integers, and
$$
x(x-a)(x-b)(x-c)+1=\left(x^{2}+d x+e\right)\left(x^{2}+f x+g\right)
$$

then let $x=0$, we get $e g=1$.
(i) If $e=g=1$, then equation (2) becomes
$$
x(x-a)(x-b)(x-c)+1=\left(x^{2}+d x+1\right)\left(x^{2}+f x+1\right)
$$

By setting $x=a, b, c$ in sequence, we obtain
$$
\left\{\begin{array}{l}
a^{2}+d a+1= \pm 1 \\
b^{2}+d b+1= \pm 1 \\
c^{2}+d c+1= \pm 1
\end{array}\right.
$$

If $a^{2}+d a+1=1$, then $a=-d$. Since $b \neq a, c \neq a$, we have
$$
\left\{\begin{array}{l}
b^{2}+d b+1=-1 \\
c^{2}+d c+1=-1
\end{array}\right.
$$

which simplifies to
$$
\left\{\begin{array}{l}b(b+d)=-2, \\ c(c+d)=-2, \\ a=-d .\end{array}\right.
$$
Solving this, we get
$$
\left\{\begin{array}{l}b=1, \\ a=3, \\ c=2 ;\end{array}\left\{\begin{array}{l}b=-1, \\ a=-3, \\ c=-2 ;\end{array}\left\{\begin{array}{l}b=2, \\ a=3, \\ c=1 ;\end{array}\left\{\begin{array}{l}b=-2, \\ a=-3, \\ c=-1 .\end{array}\right.\right.\right.\right.
$$
Thus, when at least one of the left sides of equation (4) is +1, we can solve for $(a, b, c)=(1,2,3)$ in any permutation, or $(a, b, c)=(-1,-2,-3)$ in any permutation.
If all the left sides of equation (4) are -1, then by subtracting the equations pairwise, we get
$$
\left\{\begin{array}{l}
a^{2}-b^{2}+d(a-b)=0, \\
b^{2}-c^{2}+d(b-c)=0, \\
c^{2}-a^{2}+d(c-a)=0 .
\end{array}\right.
$$

Since $a-b \neq 0, b-c \neq 0, c-a \neq 0$, we have
$$
\left\{\begin{array}{l}
a+b=-d, \\
b+c=-d, \\
c+a=-d .
\end{array}\right.
$$

Thus, $a=b=c=-\frac{d}{2}$, which contradicts the fact that $a, b, c$ are distinct.
(ii) If $e=g=-1$, then equation (2) becomes
$$
x(x-a)(x-b)(x-c)+1=\left(x^{2}+d x-1\right)\left(x^{2}+f x-1\right)
$$

By setting $x=a, b, c$ in sequence, we obtain
$$
\left\{\begin{array}{l}
a^{2}+d a-1= \pm 1, \\
b^{2}+d b-1= \pm 1, \\
c^{2}+d c-1= \pm 1 .
\end{array}\right.
$$

Similarly, we can solve for
$(a, b, c)=(1,-1,2)$ in any permutation, or $(a, b, c)=(1,-1,-2)$ in any permutation.
Upon verification, we have
$$
\begin{array}{l}
x(x-1)(x-2)(x-3)+1=\left(x^{2}-3 x+1\right)^{2}, \\
x(x+1)(x+2)(x+3)+1=\left(x^{2}+3 x+1\right)^{2}, \\
x(x-1)(x+1)(x-2)+1=\left(x^{2}-x-1\right)^{2}, \\
x(x-1)(x+1)(x+2)+1=\left(x^{2}+x-1\right)^{2} .
\end{array}
$$

Therefore, the required $(a, b, c)$ are $(1,2,3),(-1,-2,-3),(1,-1,2),(1,-1,-2)$ in any permutation.