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Let $P\in{\mathbb Z}[X]$ be a monic polynomial of degree $d>1$.

When $d=2$ or $3$, it is easy to see that at least one of $P-1,P-2,\ldots,P-(d+1)$ is irreducible over $\mathbb Q$ (see below). Does this property still hold for $d \geq 4$ ?

Explanation for the $d\leq 3$ case : in this case, being reducible is equivalent to having a root. So $P-1$ must have a root $i_1\in{\mathbb Z}$ and $P-2$ must have a root $i_2\in{\mathbb Z}$. We can write $P-1=(X-i_1)Q(X-i_1)$ where $Q$ is a polynomial of degree $d-1$. We may assume without loss of generality that $i_1=0$. Then $P-1=XQ(X)$ $1=P(i_2)-1=i_2Q(i_2)$. So the integer $i_2$ is a divisor of $1$ ; we have $i_2\in \lbrace \pm 1\rbrace$. Replacing $P(X)$ with $P(-X)$, we may assume $i_2=1$. Then $Q(1)=1$, so $Q=1+(X-1)R$ where $R$ is a polynomial of degree $d-2$. This yields $P=1+X(1+(X-1)R)$. When $d=2$, we must have $R=1$ because $P$ is monic. Similar consideration finish off the $d=3$ case.

Ewan Delanoy
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Let $P(z)=Az(z-1)(z-2)(z-3)+z+1$. You get four reducible polynomials in a row; can you choose $A$ so that a fifth is the product of two quadratics?

Let $B=1/A$, $BP(z)=z^4-6z^3+11z^2-6z+B(z+1)$. Suppose this is $(z^2-6z+C)(z^2+D)$ I need to solve these equations: $$C+D=11,-6D=B-6,CD=B$$ This becomes $D^2-17D+6=0$. Solve that quadratic, and you have $D,C,B$ and $A$.
So it is possible to have five reducible quartics in a row.

Empy2
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    The existence of such an $A$ (after some algebra) is equivalent to the existence of two integers $b,c$ with $-4b^3 + (-c - 24)b^2 + (2c - 44)b + (c^2 + 13c - 24)=0$ (this condition is necessary and sufficient for $z^2+bz+c$ to divide $P(z)$). I found no solution so far – Ewan Delanoy Oct 31 '15 at 16:33
  • Whoops, I didn't notice the $\mathbb{Q}$ condition. – Empy2 Oct 31 '15 at 16:40
  • That's what I was going to say! Your $D$ is not rational. – Ewan Delanoy Oct 31 '15 at 16:40
  • Your condition is a quadratic in $c$; what is the discriminant? – Empy2 Oct 31 '15 at 16:41
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    The discriminant is $ b^4 + 12b^3 + 74b^2 + 228b + 265=(b^2+6b+19)^2-96$ (irreducible in $b$), this suggests that $c$ should be near $b^2+6b+19$. – Ewan Delanoy Oct 31 '15 at 16:47
  • Does $\Delta=(b+3)^4+20(b+3)^2+4=(b^2+6b+19)^2-96$ help at all? – Empy2 Oct 31 '15 at 16:50
  • Oh, I see : $\Delta$ and $\Delta+96$ must both be perfect squares, and there are only finitely many solutions : $\Delta$ must be one of $2^2,5^2,10^2,23^2$. – Ewan Delanoy Oct 31 '15 at 16:53
  • Thid yields five possibilities for $b$ : $b\in\lbrace -1,-2,-3,-4,-5\rbrace$. For each value of $b$, there are two corresponding roots for the quadratic in $c$. I checked that none of them work. So, we have proved that this particular idea will not work. – Ewan Delanoy Oct 31 '15 at 17:19