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.
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.