Divisive succession

Note that for any polynomial of degree at most $3$ , and any value of $x$ , $f(x) - 4f(x+1) + 6f(x+2) - 4f(x+3) + f(x+4) = 0$ . It is easy to prove this - for example, notice that it is the method of differences (from a Brilliant post a while back) applied iteratively four times.

Let us suppose that $n$ is at least $7$ . Then, if $2 \leq x \leq 4$ , $f(x) | f(x+1)$ , $f(x) | f(x+2)$ and $f(x) | f(x+3)$ . But $f(x+3) = 4f(x+2) - 6f(x+1) + 4f(x) - f(x-1)$ , so $f(x) | f(x-1)$ (since $f(x)$ divides all the other terms in that expression. Hence, $|f(x)|$ is constant for $1 \leq x \leq 4$ .

Let $f(1) = A$ . There are 8 cases for the values of $f(2)$ , $f(3)$ and $f(4)$ , since each must be $\pm A$ .

I will not deal with all 8 cases here, but I will deal with three as examples.

1. $A, A, A, A, A, A...$ does not work, since $f$ is constant.
2. $A, -A, -A, A, 5A, 11A$ does not work, since $5 \nmid 11$ .
3. $A, -A, -A, -A, -3A, -9A, -21A$ does not work, since $9 \nmid 21$ .

We have worked out subsequent values of $f$ using the above formula.

Dealing with the other five cases similarly shows that $n < 7$ .

The third case gives us our example when $n = 6$ : Taking $A = -3$ , we find that $f(x) = (x-2)(x-3)(x-4)+3$ works.