2

My professor proposed me this question.

Suppose we have a polynomial $A(x)$ with integer coefficients. This polynomial is special in that for all such $x, y$ integers, $A(x)$ divides $A(x+y)-A(y)$. What are all the possible polynomials $A(x)$?

All I have so far is that all constant functions and all linear functions work.

user283985
  • 51

1 Answers1

1

It suffices to use information gotten from $y=1$ alone.

Extended hints (assuming $A(x)$ is a non-constant polynomial of the prescribed type):

  1. Show that $D(x):=A(x+1)-A(1)$ shares the degree and the leading coefficient with $A(x)$.
  2. Show that $$\lim_{x\to\infty}\frac{D(x)}{A(x)}=1.$$
  3. Conclude that we must have $D(x)=A(x)$ for all large enough integers $x$.
  4. Conclude that $D(x)=A(x)$ for all integers $x$.
  5. Show that $A(x)$ must be linear (with zero intercept).
Jyrki Lahtonen
  • 133,153
  • 1: A(1) is a constant so it must have same degree, 2 is true by l'hopitals, 3 is true due to the limit, not sure how to show 5 yet – user283985 Oct 26 '15 at 07:52
  • Correct. Initially step 1 was created with $D(x)=A(x+1)-A(x)$ in mind (when $D(x)$ would have strictly lower degree), but then I realized that I needed this form instead (me thinks). Sorry about leaving such a trivial step. – Jyrki Lahtonen Oct 26 '15 at 07:56
  • I see now the proof conceptually, but I'm struggling to form it into words. Could I see a sample proof? – user283985 Oct 26 '15 at 07:57
  • If $A(x)$ is of degree $n>0$, then in step 5 you can compare the coefficients of $x^{n-1}$-term of both $D(x)$ and $A(x)$ to conclude that we must have $n=1$. I've got to commute now. Looks like you're well on your way to a solution. Good luck with the rest! – Jyrki Lahtonen Oct 26 '15 at 08:00
  • Ah, I see. Thanks! – user283985 Oct 26 '15 at 08:03