Unique Quadratic
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 .
The answer is 5.
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2 solutions
This solution is basically correct, but is written very sketchily, a lot of details are missing. Please remember: to be considered complete, a solution should be detailed enough to convince your peers, who have not solved the problem, that your answer is correct. It is not enough to convince the grader that you solved it. We realize that writing complete solutions takes a lot of time and effort, but we feel that this is a valuable skill to acquire and polish, so this effort is well spent. Indeed, every successful researcher, in mathematics and other sciences, must put a lot of effort in explaining his or her results to the scientific community. Naturally, these explanations must be targeted at the audience, whose understanding of the matter at hand is comparable to yours or slightly inferior.
First, note that for f ( x ) = x 2 − 5 x + 5 one can choose n = 5 . We will now prove that n = 5 is the largest possible.
If f ( 1 ) = 0 , then f ( 2 ) = … = f ( n ) = 0 , so, since f is non-constant, n ≤ 2 . If f ( 1 ) = 0 , divide all coefficients of f by f ( 1 ) . This way we will get a polynomial with rational coefficients, that takes integer values at 1 , 2 , … , n with f ( 1 ) = 1 and f ( i ) ∣ f ( i + 1 ) for i = 1 , … , n − 1 . We will replace the original polynomial by this polynomial, which we will now call f .
Suppose f ( 2 ) = a , f ( 3 ) = a b , f ( 4 ) = a b c for some integers a , b and c . Then, because the polynomial f is quadratic, the difference f ( x + 1 ) − f ( x ) is linear. Because f ( 2 ) − f ( 1 ) = a − 1 and f ( 3 ) − f ( 2 ) = a b − a , we must have f ( 4 ) − f ( 3 ) = 2 ( a b − a ) − ( a − 1 ) = 2 a b − 3 a + 1 . So a b c = f ( 4 ) = a b + ( 2 a b − 3 a + 1 ) = 3 a b − 3 a + 1 .
The above identity implies that a ∣ 1 , so a = ± 1 . We will consider these two cases separately.
Case 1. a = 1 . Then we have b c = f ( 4 ) = 3 b − 2 , so b ∣ 2 , thus b can only be − 2 , − 1 , 1 , or 2 . Note that a and b completely determine f , in particular f ( 5 ) = 6 a b − 8 a + 3 and f ( 6 ) = 1 0 a b − 1 5 a + 6 (either using polynomial interpolation formula or the method of differences). For a = 1 this means f ( 5 ) = 6 b − 5 and f ( 6 ) = 1 0 b − 9 . If b = − 2 , then f ( 4 ) = − 8 and f ( 5 ) = − 1 7 , so n ≤ 4 . If b = − 1 , then f ( 4 ) = − 5 and f ( 5 ) = − 1 1 , so n ≤ 4 . If b = 1 , then f must be the constant polynomial 1 , which is not allowed. If b = 2 , then f ( 4 ) = 4 and f ( 5 ) = 7 , so, again, n ≤ 4 .
Case 2. a = − 1 . Similarly to case 1, we get − b c = f ( 4 ) = − 3 b + 4 , so b ∣ 4 . We have in this case f ( 5 ) = − 6 b + 1 1 , f ( 6 ) = − 1 0 b + 2 1 . Because f ( 5 ) is odd, if f ( 4 ) divides f ( 5 ) , it must be odd, so b must be odd. Thus, we just need to check b = − 1 and b = 1 . If b = − 1 , then f ( 4 ) = 7 and f ( 5 ) = 1 7 , so n ≤ 4 . If b = 1 , then f ( 4 ) = 1 , f ( 5 ) = 5 , f ( 6 ) = 1 1 . So in this case n ≤ 5 and n = 5 can be achieved: this happens for f ( x ) = x 2 − 5 x + 5 , or any multiple of it.
So, with all the cases considered, the largest possible value of n is 5 .
Let f(1), f(2), f(3), f(4) be a,xa,ya,za.
Then z-3y+3x-1=0. Since x|y, x|z, x must be 1 or -1
(1)If x=1, then z=3y-2, f(5)=(6y-5)a, f(4)|f(5) => (3y-2)|1 => y=1 contradiction
(2)If x=-1, then z=3y+4, f(5)=(6y+11)a, f(4)|f(5) => (3y+4)|3 =>y=-1, f(1)=a,f(2)=-a,f(3)=-a,f(4)=a,f(5)=5a,f(6)=11a
so max(n)=5