Polynomials $f$ and $f'$ with all roots distinct integers Edit 2. Since the question below appears to be open for degree seven and above, I have re-tagged appropriately, and also suggested this on MathOverflow (link) as a potential polymath project.
Edit 1. A re-phrasing thanks to a comment below:

Is it true that, for all $n \in \mathbb{N}$, there exists a degree $n$ polynomial $f \in \mathbb{Z}[x]$ such that both $f$ and $f'$ have all of their roots being distinct integers? (If not, what is the minimal $n$ to serve as a counterexample?)

The worked example below for $n = 3$ uses $f$ with roots $\{-9, 0, 24\}$ and $f'$ with roots $\{-18, -4\}$.
(See also the note at the end, and the linked arXiv paper.)


Question. For all $n \in \mathbb{N}$: Is it possible to find a polynomial in $\mathbb{Z}[x]$ with $n$ distinct $x$-intercepts, and all of its turning points, at lattice points?

This is clearly true when $n = 1$ and $n = 2$. A bit of investigation around $n = 3$ leads to, e.g., the polynomial defined by:
$$f(x) = x^3 + 33x^2 + 216x = x(x+9)(x+24)$$
which has $x$-intercepts at $(0,0)$, $(-9, 0)$, and $(-24, 0)$. Taking the derivative, we find that:
$$f'(x) = 3x^2 + 66x + 216 = 3(x+4)(x+18)$$
so that the turning points of $f$ occur at $(-4, -400)$ and $(-18, 972)$.
I am not even sure if this is true in the quartic${^1}$ case; nevertheless, this question concerns the more general setting. In particular, is the statement true for all $n \in \mathbb{N}$ and if not, then what is the minimal $n$ for which this is not possible?

$1$. Will Jagy kindly resolves $n=4$ since the monic quartic $f$ with integer roots $\{-7, -1, 1, 7\}$ leads to an $f'$ with roots $\{-5, 0, 5\}$. This example is also found as B5 in the paper here (PDF 22/24). The same paper has the cubic example above as B1, and includes a quintic example as B7:
$$f(x) = x(x-180)(x-285)(x-460)(x-780)$$ 
$$\text{ and }$$
$$f'(x) = 5(x-60)(x-230)(x-390)(x-684)$$
The linked arXiv (unpublished) manuscript seems to suggest that this problem is open.
 A: The arXiv paper posted here (pdf) contains a list of references that have broached this problem in the past, and examples for $n=3$, $4$, and $5$ (which I have since incorporated into the OP). 
However, even the case of $n = 6$ is listed as open$^\star$ (cf. Open Problem 6 on PDF 23/24) as of 2004. So, it appears the question asked here is open.

$\star$ (Edit): User i9Fn helpfully points to a 2015 paper containing the sextic polynomial
$$f(x) = (x − 3130)(x + 3130)(x − 3590)(x + 3590)(x − 10322)(x + 10322)$$
which leads to 
$$f'(x) = 6x(x − 3366)(x + 3366)(x − 8650)(x + 8650)$$
thereby resolving the above open problem, and leading to an updated open question (cf. p. 363):

Are there polynomials with the above-stated features and degree greater than six?

According to this latter paper's author, no such examples are known.
A: $$ x^4 - 50 x^2 + 49  = (x-1)(x+1)(x-7)(x+7) $$
$$  4 x^3 - 100 x = 4x (x-5)(x+5) $$
A: Generalizing Will Jagy's quartic solution as follows.
If $a$ and $n$ satisfy the Pell equation
$$
a^2+1=2n^2,
$$
then the quartic
$$P(x)=(x^2-1)(x^2-a^2)$$
works as its derivative is
$$
P'(x)=4x(x^2-n^2).
$$
Solutions to this Pell equation are found as follows. Let
$$
(1+\sqrt2)^{2k+1}=A_k+N_k\sqrt2.
$$
Then
$$
A_k^2-2N_k^2=(A_k+N_k\sqrt2)(A_k-N_k\sqrt2)=(1+\sqrt2)^{2k+1}(1-\sqrt2)^{2k+1}=(-1)^{2k+1}=-1,
$$
so $a=A_k$, $n=N_k$ is a solution.
With $k=1$ we get $A_1=7$, $N_1=5$ and Will's polynomial
$$
P(x)=(x+7)(x+1)(x-1)(x-7)
$$
with derivative
$$
P'(x)=4(x+5)x(x-5).
$$
Similarly, with $k=2$ we arrive at $A_2=41$, $N_2=29$ and the polynomial $P(x)=(x^2-1)(x^2-41^2)$
with extrema at $0$ and $\pm 29$.
A: Try working backwards: find an integer polynomial $F$ of degree $n-1$ with all integer roots, such that its antiderivative has $n$ distinct roots. One way to check for this by looking for $n$ sign changes. Here's an example for $n-4$, I'll edit when I find a general solution.
