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Now we have a Polynomial $P(n)$ on $\mathbb{Z}[x]$. It can't be wriiten as $P(n)=F(n)G(n)$ while $F(n),G(n) \neq 1$. Is it right that for any $P(n)$, there is a $n$ such that $P(n)$ is a prime? Is it right that for any $P(n)$, there is a $n$ such that $P(n)$ is not a square-number?

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$P(n)=n^2+n+4$ is a counterexample. – Dag Oskar Madsen Oct 28 '12 at 14:11
Thanks. I've modified it. – Lwins.Gafield Oct 28 '12 at 14:22
This is classical. See – François Brunault Oct 28 '12 at 14:54
Please explain the motivation for your question. Do you mean $\not= \pm 1$ when you write $\not=1$? I assume that by "prime" you mean "positive or negative prime". – Goldstern Oct 28 '12 at 14:57
The second part is a special case of Hilbert's irreducibility theorem; indeed it's necessary and sufficient that $P$ is not the square of another polynomial. This guarantees that the $k$-th finite difference of $\sqrt{P(n)}$ is not identically zero for any $k$. But once $k > \frac12 \deg(P)$, that finite difference approaches zero as $n \rightarrow \infty$, so once $m$ is large enough one of $\sqrt{P(m+1)}$, $\sqrt{P(m+2)}$, $\sqrt{P(m+3)}$, ..., $\sqrt{P(m+k)}$ is not an integer. – Noam D. Elkies Oct 28 '12 at 14:58

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