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Let $ P \in \mathbb{Z}[X] $ monic of degree d such that there exists an infinite sequence $ (x_i) \subset \mathbb{Z} $ , where $ |P(x_i)| $ is prime or equal to 1.

Show that P is irreducible and that $ 2d+1$ integers $ x_i$ are sufficient to prove this.

Psylex
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Suppose $P$ is reducible i.e. suppose $P(x)=f(x)g(x)$ where $f,g \in \mathbb{Z}[X]$.

If $\text{dim}(f)=m$ and $\text{dim}(g)=n$ then we have $1 \le m \lt d$, $1 \le n \lt d$ and $m+n=d$.

We know that $|P(x_i)|$ is prime or $1$ at each $x_i$. But since $f(x_i)$ and $g(x_i)$ are integers then we must have $f(x_i)= \pm 1$ or $g(x_i)= \pm 1$ (or both) at each $x_i$.

But there are at most $2\text{dim}(f)=2m$ roots of $f(x)= \pm1$ ...

gandalf61
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    Same classic proof in Mott & Rose that I linked here. Anyone interested in this and related problems should read that nice little survey. – Bill Dubuque Feb 06 '19 at 17:29
  • Hence $ f(x) \pm 1 $ has infinitely many roots so it's the null polynomial and so either f or g are units if I am not mistaken.

    Decent magic you got there gandalf61 cheers.

     – Psylex Feb 06 '19 at 17:45