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Suppose $f$ is a uni-variate polynomial of degree at most $2k-1$ for some integer $k\geq1$. Let $f^{(m)}$ denote the $m$-th derivative of $f$. If $f$ and $f^{(m)}$ have $k$ distinct common roots then, Is it true that $f$ has to be a zero polynomial? Here $m<k$ is a positive integer. This statement is true for $m=1$ but is it true for larger $m$ also?

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2 Answers 2

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Assume $a,b,c \in \mathbb{R}$ solve $$2(a^3+b^3+c^3)-3(a^2b+ab^2+b^2c+bc^2+a^2c+ac^2)+12abc=0,$$ e.g. $(a,b,c)=(-1,1,3)$. Then $$ \begin{eqnarray} f(x)&:=&(x-a)(x-b)(x-c)(3x^2-2(a+b+c)x+3(ab+bc+ca)-2(a^2+b^2+c^2))\\ &=&3x^5-5(a+b+c)x^4+10(ab+bc+ca)x^3\\ &&+(2(a^3+b^3+c^3)-3(a^2b+ab^2+\dots)-18abc)x^2+\dots \end{eqnarray} $$ satisfies $$f^{(2)}(x)=60(x-a)(x-b)(x-c)$$ and is therefore a counterexample for $m=2$ and $k=3$.

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  • $\begingroup$ Yeah, this counterexample seems to work. Do you think that such counterexamples can always be constructed for larger $k$ also? The problem I was working on cares only whether this statement is true for large $k$. Namely whether is it true for $\forall k\geq k_0$ for some $k_0$? $\endgroup$ Apr 22, 2015 at 13:51
  • $\begingroup$ What is the explicit counterexample? Fix a,b and solve the cubic for c. $\endgroup$
    – joro
    Apr 22, 2015 at 14:14
  • $\begingroup$ The explizit counterexample is $f(x)=(x^3-3x^2-x+3)(3x^2-6x-25)$. $\endgroup$
    – user35593
    Apr 22, 2015 at 14:50
  • $\begingroup$ You could try a similar ansatz for $k>3$, i.e. $f(x)=(x-a_1)\dots(x-a_k)(x^2+b_1x+b_0)$. You then get $k$ equations for the $k+2$ unknowns $a_1,\dots,a_k,b_1,b_0$. However I dont know if these equations have a solution. $\endgroup$
    – user35593
    Apr 22, 2015 at 15:23
  • $\begingroup$ By the way the LHS above is equal to $(2a-b-c)(2b-a-c)(2c-a-b)$ so the condition is satisfied iff $a,b,c$ are in arithmetic progression. $\endgroup$
    – user35593
    Apr 22, 2015 at 15:52
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It's plausible that the answer is yes. Namely, let $q(x) = \prod (x-r_i)$ over the common roots $r_i.$ Then, we want

$$p^{(m)}(x) = q(x) r(x),$$ $$p(x) = q(x) s(x).$$

On the other hand, $p(x)$ is the $m$-fold integral of $p^{(m)}$ plus a polynomial of degree $m-1.$ Unravelling all this, we get a system with $2k-1$ (linear) equations with $2k-2$ unknowns, so in general it should have no solutions (but it is possible that some choices of $q(x)$ will be exceptional).

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