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The linked arXiv (unpublished) manuscript seems to suggest that this problem is open.

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  • $\begingroup$ So you meant both $f',f$ are integer polynomials with distinct integer roots $\endgroup$ Commented Jun 6, 2017 at 20:42
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    $\begingroup$ We can rephrase it as : $f',f$ are rational polynomials with distinct and all rational roots. We can probably replace $D : f \mapsto f'$ by any $\mathbb{Q}$ linear operator $T : \mathbb{Q}[X] \to \mathbb{Q}[X] $, the situation shouldn't be very different. And we can investigate this $\bmod p$ for each $p$. $\endgroup$ Commented Jun 6, 2017 at 21:05
  • $\begingroup$ @user1952009 Thanks for the assistance with re-phrasing. If you have ideas about how to broach the problem, I would welcome them in the form of an answer! $\endgroup$ Commented Jun 6, 2017 at 22:59
  • $\begingroup$ This theorem might help to prove $f'$ or $f$ has no rational roots. And see the references of this article they are about $f,f'$ having rational or integer roots $\endgroup$ Commented Jun 7, 2017 at 0:10
  • $\begingroup$ @user1952009 Thanks: The latter article references an AMM piece, which I looked up to see what had cited it. In this way, I came upon this... $\endgroup$ Commented Jun 7, 2017 at 0:19

5 Answers 5

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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.

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    $\begingroup$ This paper claims to found infinitely many sextic polynomials.This also looks interesting, but I can't read them. $\endgroup$ Commented Jun 10, 2017 at 6:36
  • $\begingroup$ @i9Fn Thanks! I located a copy of the paper and have updated accordingly. $\endgroup$ Commented Jun 15, 2017 at 16:37
  • $\begingroup$ I could not find this thread, so temporarily reasked. But one of the viewers did find this thread, so I deleted that version. Recording the following linke found by Sil in case it is not among those listed here. $\endgroup$ Commented Nov 19, 2022 at 4:44
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    $\begingroup$ @JyrkiLahtonen I see a PDF to the Groves article. Thanks for the link; RIP to the author 🕊️ $\endgroup$ Commented Nov 19, 2022 at 14:18
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$$ x^4 - 50 x^2 + 49 = (x-1)(x+1)(x-7)(x+7) $$ $$ 4 x^3 - 100 x = 4x (x-5)(x+5) $$

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    $\begingroup$ How did you find it ? Randomly or is there an idea behind this ? $\endgroup$ Commented Jun 7, 2017 at 0:00
  • $\begingroup$ @reuns A Pell equation lies under this construction. See my recent reincarnation. $\endgroup$ Commented Nov 18, 2022 at 21:41
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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 $\pm 29$.

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  • $\begingroup$ Very cool! Thanks (: $\endgroup$ Commented Nov 18, 2022 at 22:25
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It is likely that this is impossible in degree six, with some very predictable behavior that might, perhaps, be provable.

First, given distinct integer roots of the sextic $f(x)$ $$ 0 < e < d < c < b < a, $$ it seems that we cannot get as many as three integer roots of the quintic $f'(x)$ unless $a$ is even, meanwhile $b+e = a$ and $c + d = a.$ Which is to say that a simple integer translate $$ f\left(x - \frac{a}{2} \right) = (x^2 - u^2)(x^2 - v^2)(x^2 - w^2). $$

====================================

     a     b     c     d     e
    16    13    11     5     3     Crit  :       1     8    15   total   3
    26    24    15    11     2     Crit  :       6    13    20   total   3
    30    23    22     8     7     Crit  :       2    15    28   total   3
    32    26    22    10     6     Crit  :       2    16    30   total   3
    38    32    28    10     6     Crit  :       8    19    30   total   3
    42    37    26    16     5     Crit  :       2    21    40   total   3
    46    45    24    22     1     Crit  :      10    23    36   total   3
    48    39    33    15     9     Crit  :       3    24    45   total   3
    52    48    30    22     4     Crit  :      12    26    40   total   3
    60    46    44    16    14     Crit  :       4    30    56   total   3
    62    44    32    30    18     Crit  :      22    31    40   total   3
    64    52    44    20    12     Crit  :       4    32    60   total   3
    70    59    46    24    11     Crit  :       4    35    66   total   3
    74    52    42    32    22     Crit  :      26    37    48   total   3
    74    63    48    26    11     Crit  :      18    37    56   total   3
    76    64    56    20    12     Crit  :      16    38    60   total   3
    78    64    44    34    14     Crit  :      22    39    56   total   3
    78    72    45    33     6     Crit  :      18    39    60   total   3
    80    65    55    25    15     Crit  :       5    40    75   total   3
    80    73    47    33     7     Crit  :       3    40    77   total   3
    82    70    44    38    12     Crit  :      22    41    60   total   3
    84    74    52    32    10     Crit  :       4    42    80   total   3
    86    66    52    34    20     Crit  :      26    43    60   total   3
    86    75    56    30    11     Crit  :      20    43    66   total   3
    90    69    66    24    21     Crit  :       6    45    84   total   3
    92    90    48    44     2     Crit  :      20    46    72   total   3
    96    78    66    30    18     Crit  :       6    48    90   total   3
    96    83    61    35    13     Crit  :       5    48    91   total   3
   104    96    60    44     8     Crit  :      24    52    80   total   3

=================

Second, once we focus on $$ (x^2 - p^2)(x^2 - q^2)(x^2 - r^2), $$ the computer thinks we can only factor the derivative when there is a repeat, either $p = q$ or $q = r.$

===============================

 for(int r = 1; r <= 50; ++r){
 for(int q = 1; q <= r; ++q){
 for(int p = 1; p <= q; ++p){
   mpz_class s2 = p * p + q * q + r * r;
   mpz_class s4 = q * q * r * r  +  r * r * p * p  + p * p * q * q;
   mpz_class d = s2 * s2 - 3 * s4;
  if( s2 % 3 == 0 && s4 % 3 == 0 && mp_SquareQ(d)  &&  mp_SquareQ(3 * s4)   )


     p     q     r
     1     1     1    s2: 3 =  3  s4: 3 =  3  d: 0 =  
     2     2     2    s2: 12 =  2^2 3  s4: 48 =  2^4 3  d: 0 =  
     3     3     3    s2: 27 =  3^3  s4: 243 =  3^5  d: 0 =  
     4     4     4    s2: 48 =  2^4 3  s4: 768 =  2^8 3  d: 0 =  
     1     5     5    +++   s2: 51 =  3 17  s4: 675 =  3^3 5^2  d: 576 =  2^6 3^2
     5     5     5    s2: 75 =  3 5^2  s4: 1875 =  3 5^4  d: 0 =  
     6     6     6    s2: 108 =  2^2 3^3  s4: 3888 =  2^4 3^5  d: 0 =  
     7     7     7    s2: 147 =  3 7^2  s4: 7203 =  3 7^4  d: 0 =  
     8     8     8    s2: 192 =  2^6 3  s4: 12288 =  2^12 3  d: 0 =  
     9     9     9    s2: 243 =  3^5  s4: 19683 =  3^9  d: 0 =  
     2    10    10    +++   s2: 204 =  2^2 3 17  s4: 10800 =  2^4 3^3 5^2  d: 9216 =  2^10 3^2
    10    10    10    s2: 300 =  2^2 3 5^2  s4: 30000 =  2^4 3 5^4  d: 0 =  
     1     1    11    +++   s2: 123 =  3 41  s4: 243 =  3^5  d: 14400 =  2^6 3^2 5^2
    11    11    11    s2: 363 =  3 11^2  s4: 43923 =  3 11^4  d: 0 =  
    12    12    12    s2: 432 =  2^4 3^3  s4: 62208 =  2^8 3^5  d: 0 =  
     5     5    13    +++   s2: 219 =  3 73  s4: 9075 =  3 5^2 11^2  d: 20736 =  2^8 3^4
    13    13    13    s2: 507 =  3 13^2  s4: 85683 =  3 13^4  d: 0 =  
    14    14    14    s2: 588 =  2^2 3 7^2  s4: 115248 =  2^4 3 7^4  d: 0 =  
     3    15    15    +++   s2: 459 =  3^3 17  s4: 54675 =  3^7 5^2  d: 46656 =  2^6 3^6
    15    15    15    s2: 675 =  3^3 5^2  s4: 151875 =  3^5 5^4  d: 0 =  
    16    16    16    s2: 768 =  2^8 3  s4: 196608 =  2^16 3  d: 0 =  
    17    17    17    s2: 867 =  3 17^2  s4: 250563 =  3 17^4  d: 0 =  
    18    18    18    s2: 972 =  2^2 3^5  s4: 314928 =  2^4 3^9  d: 0 =  
     1    19    19    +++   s2: 723 =  3 241  s4: 131043 =  3 11^2 19^2  d: 129600 =  2^6 3^4 5^2
    19    19    19    s2: 1083 =  3 19^2  s4: 390963 =  3 19^4  d: 0 =  
     4    20    20    +++   s2: 816 =  2^4 3 17  s4: 172800 =  2^8 3^3 5^2  d: 147456 =  2^14 3^2
    20    20    20    s2: 1200 =  2^4 3 5^2  s4: 480000 =  2^8 3 5^4  d: 0 =  
    21    21    21    s2: 1323 =  3^3 7^2  s4: 583443 =  3^5 7^4  d: 0 =  
     2     2    22    +++   s2: 492 =  2^2 3 41  s4: 3888 =  2^4 3^5  d: 230400 =  2^10 3^2 5^2
    22    22    22    s2: 1452 =  2^2 3 11^2  s4: 702768 =  2^4 3 11^4  d: 0 =  
     5     5    23    +++   s2: 579 =  3 193  s4: 27075 =  3 5^2 19^2  d: 254016 =  2^6 3^4 7^2
    13    23    23    +++   s2: 1227 =  3 409  s4: 458643 =  3 17^2 23^2  d: 129600 =  2^6 3^4 5^2
    23    23    23    s2: 1587 =  3 23^2  s4: 839523 =  3 23^4  d: 0 =  
    24    24    24    s2: 1728 =  2^6 3^3  s4: 995328 =  2^12 3^5  d: 0 =  
     5    25    25    +++   s2: 1275 =  3 5^2 17  s4: 421875 =  3^3 5^6  d: 360000 =  2^6 3^2 5^4
    11    25    25    +++   s2: 1371 =  3 457  s4: 541875 =  3 5^4 17^2  d: 254016 =  2^6 3^4 7^2
    25    25    25    s2: 1875 =  3 5^4  s4: 1171875 =  3 5^8  d: 0 =  
    10    10    26    +++   s2: 876 =  2^2 3 73  s4: 145200 =  2^4 3 5^2 11^2  d: 331776 =  2^12 3^4
    26    26    26    s2: 2028 =  2^2 3 13^2  s4: 1370928 =  2^4 3 13^4  d: 0 =  
    27    27    27    s2: 2187 =  3^7  s4: 1594323 =  3^13  d: 0 =  
    28    28    28    s2: 2352 =  2^4 3 7^2  s4: 1843968 =  2^8 3 7^4  d: 0 =  
    11    29    29    +++   s2: 1803 =  3 601  s4: 910803 =  3 19^2 29^2  d: 518400 =  2^8 3^4 5^2
    29    29    29    s2: 2523 =  3 29^2  s4: 2121843 =  3 29^4  d: 0 =  
     p     q     r

===============================

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  • $\begingroup$ Unless there is a mistake in my code there is no polynomial when the difference between the largest and smallest root is less than 200 and when it is symmetric (like you did) when the difference is less than 1000. Currently our best bet is to see if for distinct roots the value is a positive square or not and then to try and prove it (as your data suggest that but we should check for larger values considering the roots of quintic example B7. I don't understand why did you check $s2$ and $s4$ are $0 \mod 3$ and $3 * s4$ a square? $\endgroup$ Commented Jun 10, 2017 at 8:59
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    $\begingroup$ This is possible in degree six: I have updated the question to include a sextic polynomial with the desired properties. $\endgroup$ Commented Jun 15, 2017 at 17:26
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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.

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    $\begingroup$ I will undo my downvote and upvote when you have a general solution; right now, it appears that there is a solution when there is not! $\endgroup$ Commented Jun 6, 2017 at 20:22

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