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If $P\in {\mathbb Z}[X]$ is a polynomial of degree $2$, then it is easy to see that for any integer $m$, at least one of the polynomials $P-(m+1),P-(m+2),P-(m+3),P-(m+4)$ is irreducible in ${\mathbb Z}[X]$. I believe a much stronger property holds, namely that for any degree $d\geq 2$ and length $l\geq 1$, there is a constant $C(d,l)$ such that for any polynomial $P\in {\mathbb Z}[X]$ of degree $d$, in any interval $I=\lbrace m+1,m+2, \ldots ,m+C(d,l) \rbrace$ we may encounter a subinterval of length $l$, $I'=\lbrace m+j+1,m+j+2, \ldots ,m+j+l \rbrace$ with $j+l \leq C(d,l)$ such that $P-a$ is irreducible in ${\mathbb Z}[X]$ for any $a\in I'$. Is that property already known to be true, and are bounds known for $C(d,l)$ ?

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It sounds good to me. Moreover, restricting P to powers of x, I suspect you will get a good approximation on C(d,l). I also suspect that something similar fails for polynomials over the rationals, and just investigating whether C(d,l) exists for R an ordered ring (and some notion of measure that allows l to make sense) is an interesting problem. Gerhard "Ask Me About System Design" Paseman, 2011.03.29 – Gerhard Paseman Mar 29 '11 at 7:24
I assume you take $P$ of degree $d$? – Laurent Moret-Bailly Mar 29 '11 at 9:43
@ Laurent : $P$ is of degree $d$ indeed. I corrected the OP, thanks. – Ewan Delanoy Mar 29 '11 at 9:56
$6x^2+7x,6x^2+7x+1$ and $6x^2+7x+2$ are all reducible. The correct statement is with four consecutive values of quadratic polynomials. – Gjergji Zaimi Mar 29 '11 at 10:29
I do not see the point of the $m$ in the problem statement. Arbitrary polynomials of a certain degree already allow arbitrary constant term. – user11235 Apr 8 '11 at 16:35

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