## Polynomials having a common root with their derivatives

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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.

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Is your "with any" a 'there exists' or a 'for all'? – Mark Jun 11 2010 at 19:24
Mark -- "any" here means "every". – algori Jun 11 2010 at 20:05
Related question: mathoverflow.net/questions/52006/… – Felipe Voloch May 14 2011 at 22:30

## 2 Answers

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

http://front.math.ucdavis.edu/0605.5090

Not sure of its current status, though.

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Thanks, Pedro! – algori Jun 11 2010 at 20:18

5
It is still open. – quim Jun 11 2010 at 21:08

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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.

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Thanks, Andrey! – algori Jun 11 2010 at 20:30
You're welcome. – Andrey Rekalo Jun 11 2010 at 20:31
For those who do not have an access to the journal, there is an AoPS discussion at artofproblemsolving.com/Forum/… which contains enough information to recover the full proof :) – fedja May 29 2011 at 0:24
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