Problem 1384

1. **Consider the polynomial $f(x)$ with integer coefficients.**
We need to show that there exists an integer $C$ such that the set $\{ n \in \mathbb{Z} : \text{ the sum of digits of } f(n) \text{ is } C \}$ is not finite.

2. Handle the case of constant polynomials.
If $f(x)$ is a constant polynomial, say $f(x) = c$ where $c$ is an integer, then the sum of the digits of $f(n)$ is simply the sum of the digits of $c$ for all $n$. Thus, the set $\{ n \in \mathbb{Z} : \text{ the sum of digits of } f(n) \text{ is } C \}$ is infinite for $C$ equal to the sum of the digits of $c$.

3. **Assume $\deg(f) \geq 1$.**
Without loss of generality, assume the leading coefficient of $f$ is positive. If not, consider $-f$ instead, which will not change the sum of digits property we are interested in.

4. **Express $f(x)$ in its general form.**
Let $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$ where $a_i \in \mathbb{Z}$.

5. **Choose a sufficiently large $m \in \mathbb{N}$.**
We can choose $m$ such that all the coefficients of $f(x + m)$ are non-negative. This is possible because shifting the polynomial horizontally by a large enough integer $m$ will ensure that the polynomial evaluated at any integer $x$ will have non-negative coefficients.

6. **Define a new polynomial $g(x) = f(x + m)$.**
Since $g(x) = f(x + m)$, the coefficients of $g(x)$ are non-negative. This transformation does not change the degree of the polynomial or the nature of its coefficients being integers.

7. **Consider the values of $g(10^k)$ for sufficiently large $k$.**
For sufficiently large $k$, the value $10^k$ will be large enough such that the polynomial $g(10^k)$ will have a predictable pattern in its digit sum. Specifically, the leading term $a_n (10^k)^n$ will dominate the value of $g(10^k)$, and the sum of the digits of $g(10^k)$ will stabilize to a constant value $C$.

8. Conclude that the set is infinite.
Since $g(10^k)$ for sufficiently large $k$ will have the same digit sum $C$, the set $\{ n \in \mathbb{Z} : \text{ the sum of digits of } f(n) \text{ is } C \}$ is infinite.

$\blacksquare$