Here is a question someone asked me a couple of years ago. I remember having spent a day or two thinking about it but did not manage to solve it. This may be an open problem, in which case I'd be interested to know the status of it.

Let $f$ be a one variable complex polynomial. Supposing $f$ has a common root with every $f^{(i)},i=1,\ldots,\deg f-1$, does it follow that $f$ is a power of a degree 1 polynomial?

upd: as pointed out by Pedro, this is indeed a conjecture (which makes me feel less badly about not being able to do it). But still the question about its status remains.

  • 1
    $\begingroup$ Is your "with any" a 'there exists' or a 'for all'? $\endgroup$
    – Mark
    Jun 11, 2010 at 19:24
  • $\begingroup$ Mark -- "any" here means "every". $\endgroup$
    – algori
    Jun 11, 2010 at 20:05
  • $\begingroup$ Related question: mathoverflow.net/questions/52006/… $\endgroup$ May 14, 2011 at 22:30
  • 1
    $\begingroup$ The case where $\mathrm{deg}(f)=p^k$ is a prime power has been solved by Hans-Christian Graf von Bothmer, Oliver Labs, Josef Schicho, Christiaan van de Woestijne in The Casas-Alvero conjecture for infinitely many degrees, Journal of Algebra 316(1), pp. 224-230 (2007). $\endgroup$
    – Watson
    Oct 29, 2017 at 14:10

2 Answers 2


That is known as the Casas-Alvero conjecture. Check this out, for instance:


Not sure of its current status, though.

  • 7
    $\begingroup$ It is still open. $\endgroup$
    – quim
    Jun 11, 2010 at 21:08
  • $\begingroup$ In June 2018 it is still open. $\endgroup$ Jun 27, 2018 at 3:57
  • $\begingroup$ Unfortunately, the link seems to be no longer working. $\endgroup$ Jul 14, 2021 at 16:02
  • 2
    $\begingroup$ fixed url, by guessing the year $\endgroup$ Jul 15, 2021 at 22:07

The strongest result in this direction that I've heard of is Sudbery's theorem (which was originally conjectured by Popoviciu and Erdös).

Theorem. Let $P(z)$ be a polynomial of degree $n\geq 2$ and let $\Pi(z)=\prod\limits_{k=0}^{n-1}P^{(k)}(z)$ where $P^{(k)}$ is the $k$th derivative of $P$. Then either $\Pi(z)$ has exactly one distinct root or $\Pi(z)$ has at least $n+1$ distinct roots.

See the original paper by Sudbery.

  • $\begingroup$ For those who do not have an access to the journal, there is an AoPS discussion at artofproblemsolving.com/Forum/viewtopic.php?f=67&t=100933 which contains enough information to recover the full proof :) $\endgroup$
    – fedja
    May 29, 2011 at 0:24
  • $\begingroup$ One may also have a look at Rahman/Schmeisser, Analytic Theory of Polynomials, Oxford University Press, 2002, section 7.1 and notes in section 7.4 $\endgroup$ May 27, 2014 at 15:35

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