For each integer , determine all infinite sequences of positive integers , , for which there exists a polynomial of the form where , , \dots, are non-negative integers, such that for every integer .
Solution
To determine all infinite sequences of positive integers for which there exists a polynomial of the form
where are non-negative integers, and satisfying the condition
for every integer , we start by examining the implications of the given functional equation.
### Analysis
1. General Formulation:
The polynomial maps to the product . This implies that must be factorizable into exactly positive integers, each of which is a term in the sequence .
2. **Behavior for Large :**
Assume the sequence is non-decreasing and let be the common difference in an arithmetic progression starting at the maximum and continuing indefinitely. This implies .
Substituting this back into the polynomial's expression gives:
By choosing , we get the simplest case, a constant sequence. In such a situation, the polynomial simplifies to , aligning with the constant sequence's characteristics.
3. Non-Decreasing Arithmetic Sequence:
For belonging to a non-decreasing arithmetic sequence, suppose the sequence has a first term and common difference . Then, explicitly:
Thus, for the consecutive terms scenario,
4. Verification:
Given that must be a polynomial with non-negative coefficients, it immediates implies that the scaling of terms remains within the confines of the polynomial expansion. More precisely, ensuring the pattern holds for the polynomial's value provides the sequence must adhere to arithmetic constraints.
### Conclusion
From the above analysis, it follows that a sequence which satisfies the given condition must naturally form a non-decreasing arithmetic sequence since it allows to generate the required product structure for tail terms. Therefore, the valid infinite sequences in this context are precisely those that are non-decreasing and arithmetic in nature.