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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2$\begingroup$ $P(n)=n^2+n+4$ is a counterexample. $\endgroup$– Dag Oskar MadsenOct 28, 2012 at 14:11
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2$\begingroup$ This is classical. See www.mast.queensu.ca/~murty/poly2.pdf $\endgroup$– François BrunaultOct 28, 2012 at 14:54
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1$\begingroup$ 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". $\endgroup$– GoldsternOct 28, 2012 at 14:57
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1$\begingroup$ 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. $\endgroup$– Noam D. ElkiesOct 28, 2012 at 14:58
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$\begingroup$ Dear François Brunault, thank you, that helps me a lot. \\ Dear Goldstern, +-1. \\ Dear Noam D. Elkies, I've been showed something interesting by you, thanks. $\endgroup$– LwinsOct 28, 2012 at 16:25
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