73*. Polynomial

$$
y=x^{3}-18 x^{2}+a x+1784
$$

takes on three consecutive integer values at three consecutive integer points (in the same order). Find these values.
$\triangle$ Let's denote three consecutive integer values of the argument as $k-1$, $k$, and $k+1$, and the corresponding values of the polynomial as $n-1$, $n$, and $n+1$. We obtain the system of equations:

$$
\left\{\begin{array}{l}
(k-1)^{3}-18(k-1)^{2}+a(k-1)+1784=n-1 \\
k^{3}-18 k^{2}+a k+1784=n \\
(k+1)^{3}+18(k+1)^{2}+a(k+1)+1784=n+1
\end{array}\right.
$$

To eliminate $a$ and $n$ from the system, we add the first and third equations and subtract twice the second. After simplifications, we get:

$$
6 k-18 \cdot 2=0, \quad k=6
$$

The system takes the following form:

$$
\left\{\begin{array}{l}
125-18 \cdot 25+5 a+1784=n-1 \\
216-18 \cdot 36+6 a+1784=n \\
343-18 \cdot 49+7 a+1784=n+1
\end{array}\right.
$$

Subtracting the second equation from the third in this system, we get:

$$
127-18 \cdot 13+a=1, \quad a=108
$$

Then

$$
n=216-18 \cdot 36+6 \cdot 108+1784=2000
$$

We need to verify the first and third equations. Please do this verification yourself.

Answer: $1999,2000,2001$.