Three. (50 points) Find all integer triples $(a, b, c)$ $(a \neq 0)$ such that $f(x)=a x^{2}+b x+c$ has the following property: For any integer $n$, there exists an integer $m$ satisfying
$$
f(m)=f(n) f(n+1) .
$$
Three, Lemma: Let $P(x)$ be a quartic polynomial with integer coefficients. If for any integer $n$, $P(n)$ is a perfect square, then $P(x)$ is a perfect square polynomial.
Proof: Let
$$
\begin{array}{l}
P(x)=A x^{4}+B x^{3}+C x^{2}+D x+E \\
=A Q^{2}(x)+R(x),
\end{array}
$$

where $Q(x)=x^{2}+\lambda x+\mu$, to ensure $\operatorname{deg} R \leqslant 1$, then $\lambda, \mu \in \mathbf{Q}$ satisfy:
$$
2 A \lambda=B, A\left(\lambda^{2}+2 \mu\right)=C .
$$

Clearly, $A>0$.
Assume $A$ is not a perfect square, then
$$
\begin{array}{l}
\lim _{n \rightarrow \infty}(\sqrt{A} Q(n)-\sqrt{P(n)}) \\
=\lim _{n \rightarrow \infty}\left(-\frac{R(n)}{\sqrt{A} Q(n)+\sqrt{P(n)}}\right)=0 .
\end{array}
$$

Now take $\varepsilon>0$, such that
$$
\{2 \sqrt{A}\} \in(4 \varepsilon, 1-4 \varepsilon) \text {. }
$$

Thus, there exists $N \in \mathbf{Z}_{+}$, when $n \geqslant N+2$, we have $|\sqrt{A} Q(n)-\sqrt{P(n)}|1-4 \varepsilon$, a contradiction.

Now take a positive integer $M$ such that $M \lambda$ and $M \mu$ are both integers, then $M \sqrt{A} Q(n)-M \sqrt{P(n)}$ is an integer.
Notice,
$$
\lim _{n \rightarrow \infty}(M \sqrt{A} Q(n)-M \sqrt{P(n)})=0 \text {. }
$$

This indicates that for sufficiently large $n$, we have
$$
\sqrt{P(n)}=\sqrt{A} Q(n) \text {. }
$$

By the polynomial identity theorem, for any $x$, we have $P(x)=A Q^{2}(x)$.
The lemma is proved.
For any integer $n$, the equation
$$
a x^{2}+b x+c-f(n) f(n+1)=0
$$

has an integer solution.
Thus, $\Delta=b^{2}-4 a c+4 a f(n) f(n+1)$ is a perfect square, which is a quartic polynomial in $n$, denoted as $g(n)$, with the leading coefficient $4 a^{3}$.

By the lemma, we know that $a$ is a perfect square, and $g(n)$ is a perfect square polynomial.
Let $a=k^{2}, g(n)=\left(2 k^{3} x^{2}+t x+s\right)^{2}$.
Comparing the coefficients of the cubic and quadratic terms, we get
$$
t=2 k\left(b+k^{2}\right), s=k(b+2 c) \text {. }
$$

Comparing the constant term, we get
$$
k^{2}(b+2 c)^{2}=b^{2}-4 c k^{2}+4 k^{2} c\left(k^{2}+b+c\right) \text {. }
$$

Simplifying, we get $\left(k^{2}-1\right)\left(b^{2}-4 c k^{2}\right)=0$.
If $k^{2}=1$ i.e., $a=1$, for each pair $(b, c)$, we can take
$$
m=n^{2}+(b+1) n+c
$$

satisfying the condition.
If $k^{2}>1$, then $f(x)$ is a perfect square polynomial.
Let $f(x)=(k x+l)^{2}$. Then
$$
(k n+l)(k n+k+l)= \pm(k m+l) \text {. }
$$

We only need $l^{2} \equiv \pm l(\bmod k)$, i.e.,
$k \mid\left(l^{2}-l\right)$ or $k \mid\left(l^{2}+l\right)$.
In summary, all the integer triples satisfying the condition are $(1, b, c)(b, c \in \mathbf{Z})$ or $\left(k^{2}, 2 k l, l^{2}\right)\left(k \in \mathbf{Z}_{+}\right.$, and $k!\left(l^{2}-l\right)$ or $\left.k!\left(l^{2}+l\right)\right)$.