Integer polynomials with translation of the constant term - Are these examples unique?

Question

Consider the polynomial $$(x+k_1)(x+k_2)(x+k_3)(x+k_4)+c \in \mathbb{Z}[x]$$ for given $c \in \mathbb{Z}$ and $k_1,k_2,k_3,k_4 \in \mathbb{Z}$.

In this question, a user is asked to find for which constants $k \in \mathbb{Z}$ the polynomial $(x+1)(x+2)(x+3)(x+k)+1$ is factorable as a square of a quadratic polynomial. As it turns out, the only values are $k = 0,4$. By translation, we can obviously find an infinite family of solutions with $(c,k_1,k_2,k_3,k_4)=(1,l,l+1,l+2,l+3)$ where $l \in \mathbb{Z}$.

Naturally, a generalization of this is the question above is: Are there any other examples (up to translation) aside from those already stated, $c = 1$ or not? As the discrimant of this polynomial vanishes iff a factor with multiplicity bigger than $1$ appears, I tried a couple of different spacings of the variables $k_i$ with one variable undeterminded, but that did not lead to another solution (maybe I haven't searched hard enough?).

Generalizing this even further: What about a different numbers of factors on the left side (with undeterminded variables defined accordingly) and an $n$-th power on the right side? Of course, considering the order of the polynomial on the left, $n$ has to be a divisor - other than 1 - of the number of factors on the right side, but I haven't investigated further in this direction. Formalized this would look like:

$$?\exists n,l \in \mathbb{N}, k_1,\ldots,k_l \in \mathbb{Z}: \left(\prod\limits_{i=1}^{l}(x+k_i)\right)+c = p(x)^n$$

where $p(x) \in \mathbb{Z}[x]$.

So, my two questions on this matter are

1. When is $(x+k_1)(x+k_2)(x+k_3)(x+k_4)+1$ factorable as a quadratic polynomial? What about $(x+k_1)(x+k_2)(x+k_3)(x+k_4)+c$ with $c \in \mathbb{Z}$?
2. As a general problem: When is $\left(\prod\limits_{i=1}^{l}(x+k_i)\right)+c$ factorable as an $n$-th power of a polynomial , where $n,l \in \mathbb{N}$?

I am happy about every thought or solution on this, even though I am most comfortable with an undergraduates course worth of Abstract Algebra.

EDIT: So I have been thinking about this problem a bit more today and actually involved Mathematica to aid in the quest to more examples. For 2., I haven't actually made any advances yet; but if I'm not mistaken, I found and amazing algebraic identity which characterizes (part of) the solution set for the simplified case (in 1.).

Sparing you of any further exposition, I present the following formula, which is independent of the integral nature of my problem:

$$\forall c, n, k \in \mathbb{C}: (x+c)(x+c+n)(x+c+n+k)(x+c+2n+k) + (\frac{n^2+nk}{2})^2 = (x^2+(2n+k+2c)x+2c^2+(k+2n)c+\frac{n^2+nk}{2})$$

Making appropriate restrictions, one arrives at a far more general solution to 1. than the one I had given. This was found with the help of Nate's answer, which pointed me to the existence of solutions iff $c$ in 1. is a square. With the power of Mathematica, I observed patterns of the differences between the terms and tried to generalize as much as possible, eventually arriving at the given formula. So far, this list seems to be exhaustive, but I have no idea on how to dis/prove the uniqueness of this solution.

Answer 1

COMMENT.-Respect of your question (1), whatever you can say about, if you want to have $(x+a)(x+b)(x+c)(x+d)+1=(x^2+Ax+B)^2$, you should have necessarily the system of four equations with six unknowns $$\begin{cases}2A=a+b+c+d\\2B+A^2=ab+ac+ad+bc+cd\\2AB=abc+abd+acd+bcd\\B^2=abcd+1\end{cases}$$ Basically the above system have infinitely many real solutions but these can be not integer ones. Besides, if you take two arbitrary values, say for $c$ and $d$, then you have as a result a determined system of four equations with four unknowns whose values can be non-real. Anyway it is an open door to possible integer solutions.

On the other hand, when $(a,b,c)=(1,2,3)$ the system above gives

$$\begin{cases}2A=6+d\\2B+A^2=11+4d\\2AB=11d+6\\B^2=6d+1\end{cases}$$ Four equations with three unknowns for which $d=0$ is compatible but this is not so for $d=4$ as you have written.

$►$ Your second question I feel it is too dreamer, I means impossible to completely solve.

Answer 2

Here's an answer to the first one:

Suppose $k_1 < k_2 < k_3 < k_4$ are integers and $$(x+k_1)(x+k_2)(x+k_3)(x+k_4)+c = (x^2+Ax+B)^2$$

Plugging in $x = -k_1$ we see that $c$ is a perfect square, let's say $c = C^2$. Moving this to the other side:

$$(x+k_1)(x+k_2)(x+k_3)(x+k_4) = (x^2+Ax+B)^2 - C^2 = (x^2+Ax+B-C)(x^2+Ax+B+C)$$

Note that the two factors on the right are both parabolas with symmetry at $x = -A/2$ so this means that one must be $(x+k_1)(x+k_4)$ and the other must be $(x+k_2)(x+k_3)$ and $k_1+k_4 = k_2+k_3 = A$. By exchanging $C$ and $-C$ wlog we get $k_1k_4 = B-C$ and $k_2k_3 = B+C$. In particular these must have the same parity.

From here we can go backwards to find solutions. Given $k_1 < k_2 < k_3 < k_4$ satisfying $k_1+k_4 = k_2+k_3$ such that $k_1k_4 - k_2k_3$ is even. Then we can take $A = k_1+k_4$, $B = \frac{k_1k_4 + k_2k_3}{2}$, and $C = \frac{k_1k_4 - k_2k_3}{2}$.