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# Problem Statement

For each integer k ≥ 2, determine all infinite sequences of positive integers a₁, a₂, ... for which there exists a polynomial P of the form:

P(x) = x^k + c_(k-1) * x^(k-1) + ... + c₁ * x + c₀

where c₀, c₁, ..., c_(k-1) are non-negative integers, such that P(aₙ) = a_(n+1) * a_(n+2) * ... * a_(n+k) for every integer n ≥ 1.

# Solution

The only infinite sequences of positive integers satisfying the given conditions are arithmetic progressions.

## Proof

We will prove this in two parts: first, we'll show that arithmetic progressions satisfy the condition, and then we'll prove that only arithmetic progressions satisfy it.

### Part 1: Arithmetic Progressions Work

If the sequence is an arithmetic progression, i.e., aₙ = d(n-1) + a₁ for some non-negative integer d, then:

a_(n+1) * a_(n+2) * ... * a_(n+k) = (aₙ + d)(aₙ + 2d)...(aₙ + kd)

So we can simply take the polynomial P(x) = (x+d)(x+2d)...(x+kd), which is of the required form.

### Part 2: Only Arithmetic Progressions Work

To prove this, we'll establish three key claims:

#### Claim 1: The sequence (aₙ) is either strictly increasing or constant.

Proof (by contradiction): Assume there's a decrease in the sequence. Then we can construct an infinite descending subsequence of positive integers, which is impossible.

#### Claim 2: There exists a constant C (depending only on the polynomial P and k) such that a_(n+1) ≤ aₙ + C.

Proof:
1. Bounding the Polynomial: Since all coefficients of P(x) are non-negative, for x ≥ 1, we have:
   P(x) ≤ x^k + c_(k-1)x^(k-1) + ... + c₁x + c₀ ≤ (1 + c_(k-1) + ... + c₁ + c₀)x^k
   Let C₁ = 1 + c_(k-1) + ... + c₁ + c₀. Then, P(x) ≤ C₁x^k for x ≥ 1.

2. Using the Given Equation: P(aₙ) = a_(n+1) * a_(n+2) * ... * a_(n+k)
   We get: a_(n+1) * a_(n+2) * ... * a_(n+k) ≤ C₁aₙ^k

3. Since the sequence is increasing (or constant), a_(n+1) ≥ aₙ, and thus:
   a_(n+1)^k ≤ a_(n+1) * a_(n+2) * ... * a_(n+k) ≤ C₁aₙ^k

4. Taking the k-th root of both sides:
   a_(n+1) ≤ C₁^(1/k) * aₙ

5. Let C = C₁^(1/k) - 1. Then we have the desired inequality:
   a_(n+1) ≤ aₙ + C

#### Claim 3: The difference a_(n+1) - aₙ must eventually become constant.

Proof (by contradiction using infinite descent):

1. Assume the differences don't become constant. Then there exist arbitrarily large values of n for which the difference increases.

2. Construct a subsequence (a_(nᵢ)) where the differences a_(nᵢ₊₁) - a_(nᵢ) are strictly increasing.

3. Consider the ratio P(a_(nᵢ₊₁)) / P(a_(nᵢ)):

   P(a_(nᵢ)) = a_(nᵢ + 1) * a_(nᵢ + 2) * ... * a_(nᵢ + k)
   P(a_(nᵢ₊₁)) = a_(nᵢ₊₁ + 1) * a_(nᵢ₊₁ + 2) * ... * a_(nᵢ₊₁ + k)

   Since the differences are increasing:
   a_(nᵢ₊₁ + j) ≥ a_(nᵢ + j) + (a_(nᵢ₊₁) - a_(nᵢ))  for j = 1, 2, ..., k

   Therefore:
   P(a_(nᵢ₊₁)) / P(a_(nᵢ)) ≥ (1 + (a_(nᵢ₊₁) - a_(nᵢ)) / a_(nᵢ + 1)) * ... * (1 + (a_(nᵢ₊₁) - a_(nᵢ)) / a_(nᵢ + k))

4. As i increases, this ratio grows arbitrarily large, contradicting Claim 2 which implies that this ratio should be bounded.

5. This contradiction proves that the differences must eventually become constant.

### Conclusion: The Sequence is an Arithmetic Progression

If the differences a_(n+1) - aₙ eventually become constant, say equal to d, then for some N:

a_(N+1) = a_N + d
a_(N+2) = a_(N+1) + d = a_N + 2d
...
aₙ = a_N + (n-N)d  for all n ≥ N

This shows that (aₙ) is an arithmetic progression for n ≥ N. Using the equation P(aₙ) = a_(n+1) * ... * a_(n+k), we can extend this backward and show that the entire sequence is an arithmetic progression.

Therefore, we have proved that the only infinite sequences of positive integers satisfying the given polynomial condition are arithmetic progressions.