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Does there exist a polynomial $P \in \mathbb{Z}[X]$ and pairwise distinct $a_1, \dots ,a_{d+1} \in \mathbb{Z}$, with $d > \deg(P)$, such that $|P(a_i)| = 1$ for all $i \in \{1, \dots, d+1\}$?

I am pretty sure that such a polynomial cannot exist. But this is just an intuition. Any ideas how to prove that this is a contradiction?

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Couldn't you take $P(X)=X$, $a_1=1$ and $a_2=-1$? –  Cam McLeman Jun 5 '12 at 14:59
    
Yes of course ... –  joachim Jun 5 '12 at 15:05
    
Maybe you want to insist on degree greater than 1? And maybe irreducible. –  Cam McLeman Jun 5 '12 at 15:08
    
$d$ is strictly bigger than the degree. I think this is stronger. –  joachim Jun 5 '12 at 15:11
    
Well, it's definitely stronger. Robert Israel's answer addresses this situation as well. –  Cam McLeman Jun 5 '12 at 15:12
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1 Answer 1

up vote 3 down vote accepted

If $B = \{i: P(a_i) = -1\}$, then $P(z)+1$ is divisible by $Q(z) = \prod_{i \in B} (z - a_i)$. We may assume $\text{card}(B) \ge (d+1)/2$ (otherwise replace $P$ by $-P$. Now if $i \in \{1,\ldots,d+1\} \backslash B$, $P(a_i) = 1$ so $Q(a_i)$ divides $2$, so $Q(a_i) \in \{-2,-1,1,2\}$. So each $a_i - a_j$ for $j \in B$ is in $\{-2,-1,1,2\}$, and at most one can be $\pm 2$. In particular, $\text{card}(B) \le 3$. Moreover, if $\text{card}(B) = 3$ there is at most one possible value for $a_i$, namely $0$. This reduces us to a small number of possible cases, which can be checked individually.

EDIT: One such case is $P(x) = 1 + (x+1)(x-2)$, where $P(x) = 1,-1,-1,1$ for $x=-1,0,1,2$, do $d = 3$ and $\deg P = 2$.

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Very nice. Adding filler.... –  Cam McLeman Jun 5 '12 at 15:41
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