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Algebra Difficulty 8.7 Shortlist Find the answer

For each integer k2k\geq 2, determine all infinite sequences of positive integers a1a_1, a2a_2, \ldots for which there exists a polynomial PP of the form P(x)=xk+ck1xk1++c1x+c0, P(x)=x^k+c_{k-1}x^{k-1}+\dots + c_1 x+c_0, where c0c_0, c1c_1, \dots, ck1c_{k-1} are non-negative integers, such that P(an)=an+1an+2an+k P(a_n)=a_{n+1}a_{n+2}\cdots a_{n+k} for every integer n1n\geq 1.

A number or a short expression. Spacing and $ signs are ignored.

Solution

To determine all infinite sequences of positive integers a1,a2, a_1, a_2, \ldots for which there exists a polynomial P P of the form

P(x)=xk+ck1xk1++c1x+c0, P(x) = x^k + c_{k-1}x^{k-1} + \dots + c_1 x + c_0,

where c0,c1,,ck1 c_0, c_1, \ldots, c_{k-1} are non-negative integers, and satisfying the condition

P(an)=an+1an+2an+k P(a_n) = a_{n+1}a_{n+2}\cdots a_{n+k}

for every integer n1 n \geq 1 , we start by examining the implications of the given functional equation.

### Analysis

1. General Formulation:

The polynomial P(x) P(x) maps an a_n to the product an+1an+2an+k a_{n+1}a_{n+2}\cdots a_{n+k} . This implies that P(an) P(a_n) must be factorizable into exactly k k positive integers, each of which is a term in the sequence {ai} \{a_i\} .

2. **Behavior for Large n n :**

Assume the sequence is non-decreasing and let a a be the common difference in an arithmetic progression starting at the maximum and continuing indefinitely. This implies an+i=an+(i1)d a_{n+i} = a_n + (i-1)d .

Substituting this back into the polynomial's expression gives:

P(an)=ank+ck1ank1++c1an+c0=(an+1)(an+2)(an+k). P(a_n) = a_n^k + c_{k-1}a_n^{k-1} + \ldots + c_1 a_n + c_0 = (a_{n+1})(a_{n+2})\cdots(a_{n+k}).

By choosing d=0 d = 0 , we get the simplest case, a constant sequence. In such a situation, the polynomial simplifies to P(x)=xk P(x) = x^k , aligning with the constant sequence's characteristics.

3. Non-Decreasing Arithmetic Sequence:

For an a_n belonging to a non-decreasing arithmetic sequence, suppose the sequence has a first term a1 a_1 and common difference d d . Then, explicitly:

an=a1+(n1)d. a_n = a_1 + (n-1)d.

Thus, for the consecutive terms scenario,

P(an)=(a1+nd)(a1+(n+1)d)(a1+(n+k1)d). P(a_n) = (a_1 + nd)(a_1 + (n+1)d)\cdots(a_1 + (n+k-1)d).

4. Verification:

Given that P(an) P(a_n) 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 {an} \{a_n\} which satisfies the given condition must naturally form a non-decreasing arithmetic sequence since it allows P(an) P(a_n) 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.

All non-decreasing arithmetic sequences of positive integers \boxed{\text{All non-decreasing arithmetic sequences of positive integers}}

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.