Problem 1392

1. Base Case:
- Consider $n = 2$.
- We need to find $x$ such that $3^2 \parallel f(x)$.
- Let $x = 1$. Then $f(1) = 1^3 + 17 = 18$.
- Clearly, $3^2 \parallel 18$ since $18 = 3^2 \cdot 2$ and $3^3 \nmid 18$.
- Also, $1 \not\equiv 0 \pmod{3}$.
- Thus, the base case holds.

2. Inductive Step:
- Assume for some $n \geq 2$, there exists $x \not\equiv 0 \pmod{3}$ such that $3^n \parallel f(x)$.
- We need to find $y \not\equiv 0 \pmod{3}$ such that $3^{n+1} \parallel f(y)$.

3. **Constructing $y$:**
- Given $3^n \parallel x^3 + 17$, we have $x^3 + 17 = k \cdot 3^n$ where $k$ is not divisible by 3.
- Consider $y = x + 3^{n-1}$.
- Then, $y^3 = (x + 3^{n-1})^3 = x^3 + 3 \cdot x^2 \cdot 3^{n-1} + 3 \cdot x \cdot (3^{n-1})^2 + (3^{n-1})^3$.
- Simplifying modulo $3^{n+1}$:
$$y^3 \equiv x^3 + 3 \cdot x^2 \cdot 3^{n-1} \pmod{3^{n+1}}$$
$$y^3 \equiv x^3 + 3^n \cdot x^2 \pmod{3^{n+1}}$$
- Since $3^n \parallel x^3 + 17$, we have:
$$x^3 + 17 \equiv 3^n \cdot k \pmod{3^{n+1}}$$
- Therefore:
$$y^3 + 17 \equiv x^3 + 3^n \cdot x^2 + 17 \equiv 3^n \cdot k + 3^n \cdot x^2 \pmod{3^{n+1}}$$
$$y^3 + 17 \equiv 3^n (k + x^2) \pmod{3^{n+1}}$$
- Since $k$ is not divisible by 3 and $x \not\equiv 0 \pmod{3}$, $k + x^2$ is not divisible by 3.
- Hence, $3^{n+1} \parallel y^3 + 17$.

4. Conclusion:
- We have constructed $y = x + 3^{n-1}$ such that $3^{n+1} \parallel f(y)$ and $y \not\equiv 0 \pmod{3}$.
- By induction, for each $n \geq 2$, there exists $x \not\equiv 0 \pmod{3}$ such that $3^n \parallel f(x)$ but $3^{n+1} \nmid f(x)$.

$\blacksquare$