(19) Given the function $f(x)=a x^{4}+b x^{3}+c x^{2}+d x$ satisfies:
(1) $a, b, c, d$ are all greater than 0;
(2) For any $x \in\{-2,-1,0,1,2\}, f(x)$ is an integer;
(3) $f(1)=1, f(5)=70$.
Determine whether, for each integer $x, f(x)$ is an integer. Provide an argument for your conclusion.
If $l$ is not parallel to $\alpha$, let $l \cap \alpha = P$, as shown in Figure 2. Then point $P$ is on both line $KL$ and line $MN$. On $l$, take point $Q$ such that
$$
\frac{AQ}{AC} = \frac{AN}{AD} = \frac{BL}{BC},
$$

Connect $LQ$ and $NQ$, then $LQ \parallel AB, NQ \parallel CD$, clearly we have
$$
\frac{DM}{MC} = \frac{AK}{KB} \Leftrightarrow \frac{MC}{DC} = \frac{KB}{AB}.
$$

And
$$
\frac{MC}{DC} = \frac{MC}{NQ} \cdot \frac{NQ}{DC} = \frac{PC}{PQ} \cdot \frac{AQ}{AC},
$$

Then
$$
\begin{aligned}
\frac{KB}{AB} = & 1 - \frac{AK}{AB} \\
= & 1 - \frac{AK}{QL} \cdot \frac{QL}{AB} \\
= & 1 - \frac{AP}{PQ} \cdot \frac{QC}{AC} \\
& = \frac{1}{PQ \cdot AC}(PQ \cdot AC - AP \cdot QC) \\
& = \frac{1}{PQ \cdot AC}[(PC + CQ) AC - (AC + CP) QC] \\
& = \frac{1}{PQ \cdot AC}(PC \cdot AC - PC \cdot QC) \\
& = \frac{PC \cdot AQ}{PQ \cdot AC} \\
= & \frac{MC}{DC}, \quad \frac{DM}{MC} = \frac{AK}{KB}.
\end{aligned}
$$
$$
\begin{array}{l}
\text { That is } \frac{DM}{MC} = \frac{AK}{KB}. \\
\text { Since } \\
f(1) = a + b + c + d, \\
f(-1) = a - b + c - d, \\
\text { we know } \\
f(1) + f(-1) = 2a + 2c, \\
f(1) - f(-1) = 2b + 2d. \\
\begin{array}{l}
f(2) = 16a + 8b + 4c + 2d, \\
f(-2) = 16a - 8b + 4c - 2d,
\end{array} \\
f(-2) = 16a - 8b + 4c - 2d, \\
f(2) + f(-2) = 32a + 8c, \\
f(2) - f(-2) = 16b + 4d. \\
32a + 8c - 4(2a + 2c) = 24a, \\
16(2a + 2c) - (32a + 8c) = 24c, \\
16b + 4d - 2(2b + 2d) = 12b, \\
8(2b + 2d) - (16b + 4d) = 12d,
\end{array}
$$

So $2a + 2c, 2b + 2d$ are both integers.
Since

We know
$\Delta$

So $32a + 8c, 16b + 4d$ are both integers. Also,
$$
\begin{array}{l}
32a + 8c - 4(2a + 2c) = 24a, \\
16(2a + 2c) - (32a + 8c) = 24c, \\
16b + 4d - 2(2b + 2d) = 12b, \\
8(2b + 2d) - (16b + 4d) = 12d,
\end{array}
$$
```
And $a, b, c, d$ are all greater than 0, so $24a, 12b, 24c, 12d$ are all positive integers.
Let $24a = k, 12b = l, 24c = m, 12d = n$, where $k, l, m, n$ are all positive
integers, then
$f(1) = a + b + c + d = \frac{k}{24} + \frac{l}{12} + \frac{m}{24} + \frac{n}{12}$.
```
So
$$
24f(1) = k + 2l + m + 2n,
$$

Similarly
$$
24f(-1) = k - 2l + m - 2n.
$$

Thus we have
$$
\begin{array}{l}
24[f(1) + f(-1)] = 2(k + m), \\
24[f(1) - f(-1)] = 4(l + n).
\end{array}
$$

That is, $k + m$ is a multiple of 12, and $l + n$ is a multiple of 6.
From $f(1) = 1$, we get
$$
k + m + 2(l + n) = 24,
$$

So
$$
k + m = 12, l + n = 6.
$$

From $f(5) = 70$, we get
$$
625a + 125b + 25c + 5d = 70.
$$

Thus
$$
\frac{125}{24}k + \frac{25}{12}l + \frac{5}{24}m + \frac{1}{12}n = 14,
$$

That is
$$
125k + 50l + 5m + 2n = 14 \times 24 = 336.
$$

Since
$$
\begin{aligned}
125k + 50l + 5m + 2n & = 120k + 5(k + m) + 50l + 2n \\
& = 120k + 5 \times 12 + 50l + 2n \\
& = 10(12k + 6 + 5l) + 2n,
\end{aligned}
$$

So the units digit of $2n$ is the units digit of 336, which is 6. Since $2n < 12$, we have $2n = 6, n = 3$. $l = 6 - n = 6 - 3 = 3$. Thus $125k + 5m = 180$, i.e., $24k + k + m = 36$ $24k = 24$, so $k = 1, m = 11$. Therefore,
$$
a = \frac{1}{24}, b = \frac{1}{4}, c = \frac{11}{24}, d = \frac{1}{4}.
$$

Thus
$$
\begin{aligned}
f(x) & = \frac{1}{24}x^4 + \frac{1}{4}x^3 + \frac{11}{24}x^2 + \frac{1}{4}x \\
& = \frac{1}{24}x(x^3 + 6x^2 + 11x + 6) \\
& = \frac{1}{24}x(x^3 + x^2 + 5x^2 + 5x + 6x + 6) \\
& = \frac{1}{24}x(x + 1)(x^2 + 5x + 6) \\
& = \frac{1}{24}x(x + 1)(x + 2)(x + 3).
\end{aligned}
$$
Car
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