Find the largest possible integer n such that there exists a non-constant quadratic polynomial f(x) with integer coefficients satisfying

f(1)∣f(2),f(2)∣f(3),…f(n−1)∣f(n).

Details and assumptions:
~For (possibly negative or zero) integers m and k the notation m∣k means that k=m⋅i for some integer i.

thx

can u explain how?

To find the largest possible integer n, we need to find a pattern in the divisibility condition f(k) | f(k+1), where f(x) is a non-constant quadratic polynomial with integer coefficients.

Let's start by considering a quadratic polynomial of the form f(x) = ax^2 + bx + c, where a, b, and c are integers.

We want to find conditions such that f(k) | f(k+1) for all k from 1 to n-1.

Substituting k into the polynomial f(x), we have:
f(k) = ak^2 + bk + c
f(k+1) = a(k+1)^2 + b(k+1) + c

The condition f(k) | f(k+1) implies that f(k+1) is divisible by f(k). In other words, there must exist an integer m such that f(k+1) = mf(k).

Substituting the expressions for f(k) and f(k+1), we have:
a(k+1)^2 + b(k+1) + c = m(ak^2 + bk + c)

Expanding and rearranging terms, we get:
ak^2 + (2a+b)k + (a+b+c) = m(ak^2 + bk + c)

Comparing coefficients, we have the following conditions:
1. a = ma
2. 2a + b = mb
3. a + b + c = mc

From condition 1, we have two cases:
Case 1: a = 0
If a = 0, the polynomial becomes f(x) = bx + c, which is a linear polynomial. But we are looking for a quadratic polynomial, so this case is not valid.

Case 2: a ≠ 0
If a ≠ 0, dividing both sides of condition 1 by a gives m = 1. This means that a = 1.

Substituting a = 1 into condition 3:
1 + b + c = mc
b + c = mc - 1

To satisfy condition 2, we can choose b = 1 and c = 0. This gives us a quadratic polynomial f(x) = x^2 + x with integer coefficients.

Now that we have found an example of a quadratic polynomial satisfying the divisibility condition, we can see the pattern that arises:
f(k) = k^2 + k
f(k+1) = (k+1)^2 + (k+1) = k^2 + 3k + 2 = (k+1)(k+2)

This shows that for f(k) = k^2 + k, the condition f(k) | f(k+1) holds for any positive integer k.

Therefore, the largest possible integer n satisfying the given condition is n = 2, as we can find a quadratic polynomial such that f(1) | f(2).

Note: This solution assumes that the polynomial f(x) is non-constant, which implies that it cannot be a linear polynomial.

To find the largest possible integer n that satisfies the given conditions, we can first observe the pattern of the divisibility relationship between f(x) values. Let's analyze the conditions step by step:

1. f(1)∣f(2): This condition implies that f(2) must be divisible by f(1). In other words, all the coefficients of f(2) must be divisible by the corresponding coefficients of f(1).

2. f(2)∣f(3): Similarly, this condition implies that f(3) must be divisible by f(2), requiring the coefficients of f(3) to be divisible by the corresponding coefficients of f(2).

We can continue this pattern for each subsequent condition:
3. f(3)∣f(4)
4. f(4)∣f(5)
...
n-2. f(n-2)∣f(n-1)
n-1. f(n-1)∣f(n)

From this pattern, we can see that for any condition f(k)∣f(k+1), it essentially means that the coefficient of the highest degree term in f(k+1) should be divisible by the corresponding coefficient in f(k).

Now, let's consider the quadratic polynomial f(x) = ax^2 + bx + c (where a, b, and c are integers).

To satisfy the conditions, we need to ensure that the highest degree coefficients follow the divisibility pattern. The highest degree term in f(1) is a, in f(2) is 4a + 2b + c, and in f(3) is 9a + 3b + c. We can observe that the difference between successive highest degree terms is (k^2 - (k-1)^2)a + k(k-1)b for k ≥ 2.

Using this pattern, we need to find the largest n such that for every k = 2 to n, the difference (k^2 - (k-1)^2)a + k(k-1)b is divisible by a. This ensures that the condition f(k)∣f(k+1) is satisfied for all k.

Now, to find the largest n, we need to find the smallest value of a that ensures the divisibility. The smallest positive value of a that can make (k^2 - (k-1)^2)a divisible by a, for any positive integer k, is a = 1. Therefore, we choose a = 1.

With this choice of a, the difference between successive highest degree terms simplifies to k(k-1)b.

To satisfy the divisibility condition, k(k-1)b must be divisible by a, which is 1. This means b can be any integer.

Therefore, the choice of the quadratic polynomial f(x) = x^2 + bx + c (where b and c are integers) satisfies the given conditions.

Now, to find the largest possible integer n, we need to determine when the divisibility pattern breaks. The divisibility pattern will break when the highest degree term becomes constant (i.e., the coefficient of x^2 is zero).

In our case, if a = 1, then the highest degree term will remain non-constant for any choice of b and c. Therefore, there is no largest possible integer n that satisfies the given conditions. The conditions can be satisfied for any positive integer n.

In summary, there is no limit to the largest possible integer n for which a non-constant quadratic polynomial f(x) with integer coefficients can satisfy the given divisibility conditions.