Problem 1525

To show that there exist infinitely many non-constant polynomials that are not $n$-good for any $n$, we need to construct a family of polynomials that fail to meet the criteria for being $n$-good for any non-negative integer $n$.

1. **Consider the polynomial $F(x) = Ax + 1$ where $A$ is a non-zero integer:**
- This polynomial is non-constant since $A \neq 0$.
- $F(0) = 1$, satisfying condition (a).
- For any positive integer $c$, $F(c) = Ac + 1$. Since $A$ and $c$ are positive, $Ac + 1 > 0$, satisfying condition (b).

2. **Analyze the number of primes generated by $F(x) = Ax + 1$:**
- By Dirichlet's theorem on arithmetic progressions, there are infinitely many primes in the sequence $Ac + 1$ where $c$ is a positive integer.
- This implies that for any $n$, there are infinitely many values of $c$ such that $F(c)$ is prime. Therefore, it is impossible for $F(x) = Ax + 1$ to have exactly $n$ values of $c$ such that $F(c)$ is prime.

3. **Construct a family of polynomials that are not $n$-good:**
- Consider the family of polynomials $F(x) = Ax + 1$ for any non-zero integer $A$.
- As shown, these polynomials cannot be $n$-good for any $n$ because they generate infinitely many primes, violating condition (c).

4. Conclusion:
- Since $A$ can be any non-zero integer, there are infinitely many such polynomials $F(x) = Ax + 1$.
- Therefore, there exist infinitely many non-constant polynomials that are not $n$-good for any $n$.

$\blacksquare$