Reducible polynomial + integer = Reducible polynomial? Reducible polynomial + integer = Reducible polynomial ?
As the title says.
More specific :

For every integer $n$, does there exist a pair of polynomials $p(x)$ and $q(x)$ such that:

*

*$p(x)$ and $q(x)$ each have integer coefficients with no common factor

*$p(x)$ and $q(x)$ are reducible polynomials

*$p(x) + n = q(x)$.


I mean integer nonlinear polynomials without a constant multiplication factor such as $(x^2 + 5)(x+1)$ ( but not $2x^2 $which has constant factor $2$).
I think an example clarifies :
$(x)(x-2) + 1 = (x-1)^2$
where $x(x-2)$ and $(x-1)^2$ are the reducible polynomials ( reducible means we can factor them over the ring of integers ) and $+1$ is the integer.
Are there other such identities for the integer $+1$ ?
Are there such identities for the integer $+2$ ? ( im aware that $x(x+1) = 0$ mod 2 for integer $x$ but I did not say $x$ is an integer )
How about the integer $+3$ ?
Are there such pairs of reducible polynomials for all integers $+n$ ? And if yes , how to find such ?
Edit : I used Owens answer to clarify my question.
Edit2 : I ask for additional solutions. Such as higher degree polynomials.
Or generalizations.
Such as $p(x) + 1 = q(x)$ and $p(x) + 2 = r(x)$.
 A: Here's what I think you're asking:

For every integer $n$, does there exist a pair of polynomials $p(x)$ and $q(x)$ such that:

*

*$p(x)$ and $q(x)$ each have integer coefficients with no common factor

*$p(x)$ and $q(x)$ are reducible polynomials

*$p(x) + n = q(x)$.


The answer is to this question is Yes.  Just consider
$$p(x) = x^2 - (n+1) x = (x)(x - (n + 1))$$
$$q(x) = x^2 - (n+1) x + n = (x - 1)(x - n).$$
This directly generalizes your example with $n=1$.
A: We will cheat. Let $a(x)$ and $b(x)$ be any two monic polynomials of degree $\ge 1$ with the following properties. (i) $a(x)$ has constant term $1$; (ii) $b(x)$ has constant term $-n$. Let $p(x)=a(x)b(x)$ and $q(x)=a(x)b(x)+n$. Then $q(x)$ has constant term $0$, and has degree $\ge 2$, so is reducible. 
A: Hint $\rm\  a\!+\!a' = b\!+\!b'\:\Rightarrow\: (x\!-\!a)(x\!-\!a')-(x\!-\!b)(x\!-\!b')\,=\, aa'\!-bb'$
A: $(x-p)(x-q)-(x-r)(x-s)=n$ is equivalent to $s=p+q-r$ and $(r-p)(r-q)=n$. That's full description of monic polynomials of degree $2$.
