Polynomials $f$ and $f'$ with all roots distinct integers

Question

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.)

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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?

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$1$. 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The linked arXiv (unpublished) manuscript seems to suggest that this problem is open.

Answer 1

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.

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$\star$ (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.

Answer 2

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.

Answer 3

$$ x^4 - 50 x^2 + 49 = (x-1)(x+1)(x-7)(x+7) $$ $$ 4 x^3 - 100 x = 4x (x-5)(x+5) $$

Answer 4

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

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

Answer 5

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 $\pm 29$.