1

I'm searching for a polynomial $f$ of degree 4 with the following property: $f$ and all its derivatives have the maximum number of integer roots.

Concretely formulated:

$$\begin{eqnarray*} f(x) & = & (x-a)(x-b)(x-c)(x-d) \\ f'(x) & = & 4(x-e)(x-f)(x-g) \\ f''(x) & = & 12(x-h)(x-i) \\ f'''(x) &= & 24(x-j) \\ \end{eqnarray*} $$

should be satisfied simultaneously with distinct integers $a,b,c,d$, distinct integers $e,f,g$, disctinct integers $h,i$ and an integer $j$.

My conjecture is that there is no such polynomial. For degree 3, there are solutions. Can anyone either prove this or find a counterexample?

Peter Taylor
  • 13,425
Peter
  • 41
  • 2
    Must the roots be distinct? If they are counted with multiplicity, $x^4$ will work. – Calvin Lin Apr 23 '13 at 14:33
  • Any real polynomial with exactly four distinct roots will have three distinct critical points. This is because polynomials are smooth. Therefore, between two zeros, the polymonial either goes up and comes back down or goes down and comes back up. – John Douma Apr 23 '13 at 14:42
  • Sorry, I didn't catch the integer part. – John Douma Apr 23 '13 at 14:49
  • Comparing the desired forms of the derivatives with the derivatives of the previous level we can establish the requirements that $a=b=c=d \pmod 4$ or (wlog) $a=b=c+2=d+2 \pmod 4$; $e=f=g \pmod 3$; and $h=i \pmod 2$. So a reasonable starting point would be a brute force check for non-trivial solutions in $\mathbb{Z}/12\mathbb{Z}[x]$. – Peter Taylor Apr 23 '13 at 15:31
  • Taking the indicated derivatives, you are asking for integer solutions to $$2j=h+i\3j=e+f+g\4j=a+b+c+d\3hi=ef+eg+fg\6hi=ab+ac+ad+bc+bd+cd\efg=abc+abd+acd+bcd$$ with $a,b,c,d$ distinct as well as $e,f,g$ and $h,i$. To get $h$ and $i$ rational we need $e^2+f^2+g^2-(ef+eg+fg)$ to be a perfect square. – Ross Millikan Apr 23 '13 at 16:31

1 Answers1

0

It looks like comlicated "solution" but: Try to express every polynomial in universal form:
$\int{(x-a)}dx=\frac{x^2}{2}-ax+b$ ; $\qquad \int{(\frac{x^2}{2}-ax+b)}dx = ...$
and solve system of equations (conditions of discriminants for real roots) to prove your conjuncture or find counterexample.

Community
  • 1
Stano
  • 158
  • 1
  • 10