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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The linked arXiv (unpublished) manuscript seems to suggest that this problem is open.