Nontrivial linear combinations of polynomial shifts do not cancel

Question

I have a proof of the following statement:

Let $P \in \mathbb{Z}[X]$ be a polynomial of degree $d\geq 1$, and $\alpha_0, \alpha_1, \ldots, \alpha_k \in \mathbb{Z}$, where $k\le d$. If $\sum_{i=0}^k \alpha_i \cdot P(x+i) = 0$, then all the $\alpha_i$'s are zero.

Our proof relies on a technical but easy rewriting, followed by an application of the rank theorem for Vandermonde matrices. I feel that an easier or known proof should be around. Any help?

Suppose $P(x)=x$, $a_0=1,a_1=-2,a_2=1$. Then we have $x-2(x+1)+(x+2)=0$, or am I misreading something? — lulu, Jan 15 '20 at 17:13
Did you possibly mean to connect $k$ to the degree of the polynomial or something like that? — lulu, Jan 15 '20 at 17:23
@lulu I did. Well, as there is an answer now, I will repost that question with the fix. — Michaël Cadilhac, Jan 16 '20 at 05:02
Editing the current question (as I just did for you) is better than making a new post with the correction. — David C. Ullrich, Jan 16 '20 at 12:41

score 1 · Answer 1 · answered Jan 15 '20 at 17:15

1

This is clearly false. For example if $P(x)=x$ then $-P(x-1)+2P(x)-P(x+1)=0$.

In fact for any $P$ of degree $d$ there is clearly a counterexample with $k=d+2$, since each $P(x-i)$ has degree $d$ and the space of all polynomials of degree no larger than $d$ has dimension $d+1<k$. (Hmm. Considering $\Bbb Q[x]$ we get rational $\alpha_i$, hence we can get integral $\alpha_i$.)

answered Jan 15 '20 at 17:15

David C. Ullrich

89,985

Thanks, very appreciated; I forgot indeed to indicate $k \leq d$, my bad. – Michaël Cadilhac Jan 16 '20 at 05:03

Nontrivial linear combinations of polynomial shifts do not cancel

1 Answers1