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.
To find the largest possible integer n that satisfies the given conditions, we need to find a non-constant quadratic polynomial f(x) with integer coefficients that satisfies the divisibility property f(k)|f(k+1) for all k from 1 to n-1.
Let's start by considering a general quadratic polynomial f(x) = ax^2 + bx + c, where a, b, and c are integer coefficients.
We know that f(1)|f(2), which means that f(2) is divisible by f(1). Let's substitute the values of f(1) and f(2) into the polynomial:
f(1) = a(1)^2 + b(1) + c = a + b + c
f(2) = a(2)^2 + b(2) + c = 4a + 2b + c
Since f(2) is divisible by f(1), we have:
4a + 2b + c = k(a + b + c), where k is an integer.
Expanding the equation, we get:
4a + 2b + c = ka + kb + kc
Now we rearrange the terms:
(4-k)a + (2-k)b + (1-k)c = 0
Since a, b, and c are integer coefficients, the left side of the equation must be divisible by the common factors of (4-k), (2-k), and (1-k). Thus, we have three cases to consider:
Case 1: k = 4
In this case, the equation becomes:
0a - 2b - 3c = 0
Simplifying further, we have:
2b + 3c = 0
To ensure we have a non-constant quadratic polynomial, we need a and b to be nonzero. Since the coefficients can take any integer value, we can assign a = 1 and b = -1, which gives us:
2(-1) + 3c = 0
-2 + 3c = 0
3c = 2
c = 2/3 = not an integer
Therefore, k = 4 does not yield a valid quadratic polynomial.
Case 2: k = 2
In this case, the equation becomes:
2a 0- b - c = 0
Simplifying further, we have:
-b - c = 0
b = -c
To ensure we have a non-constant quadratic polynomial, we need a and b to be nonzero. We can assign a = 1 and b = -1, so:
-b - c = 0
1 - c = 0
c = 1
Therefore, for k = 2, we have a valid quadratic polynomial f(x) = x^2 - x + 1.
Now, we need to determine the largest possible integer n for which f(k) divides f(k+1) for all k from 1 to n-1, given that f(x) = x^2 - x + 1.
We can calculate the values of f(k) and f(k+1) for various values of k:
f(1) = 1^2 - 1 + 1 = 1
f(2) = 2^2 - 2 + 1 = 3
f(3) = 3^2 - 3 + 1 = 7
f(4) = 4^2 - 4 + 1 = 13
f(5) = 5^2 - 5 + 1 = 21
From the values calculated above, we can see that f(k) divides f(k+1) for k = 1, 2, 3, and 4. However, f(4) = 13 does not divide f(5) = 21.
Therefore, the largest possible integer n that satisfies the given conditions is n = 4.