Can a Polynomial Have More Roots Than Its Degree?

[Ackbach]

Here is this week's POTW:

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Let $P(x)$ be a polynomial of degree $n$ such that $P(x)=Q(x)P''(x)$, where $Q(x)$ is a quadratic polynomial and $P''(x)$ is the second derivative of $P(x)$. Show that if $P(x)$ has at least two distinct roots then it must have $n$ distinct roots.

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[Ackbach]

Re: Problem Of The Week # 255 - Mar 11, 2017

This was Problem B-2 in the 1999 William Lowell Putnam Mathematical Competition.

No one solved this week's POTW, though there is an honorable mention to Kiwi. The solution, attributed to Kiran Kedlaya and his associates, follows:

Spoiler

Suppose that $P$ does not have $n$ distinct roots; then it has a root of multiplicity at least $2$, which we may assume is $x=0$ without loss of generality. Let $x^k$ be the greatest power of $x$ dividing $P(x)$, so that $P(x) = x^k R(x)$ with $R(0) \neq 0$; a simple computation yields
\[
P''(x) = (k^2-k)x^{k-2} R(x) + 2kx^{k-1} R'(x) + x^k R''(x).
\]
Since $R(0) \neq 0$ and $k\geq 2$, we conclude that the greatest power of $x$ dividing $P''(x)$ is $x^{k-2}$. But $P(x) = Q(x) P''(x)$, and so $x^2$ divides $Q(x)$. We deduce (since $Q$ is quadratic) that $Q(x)$ is a constant $C$ times $x^2$; in fact, $C=1/(n(n-1))$ by inspection of the leading-degree terms of $P(x)$ and $P''(x)$.

Now if $P(x) = \sum_{j=0}^n a_j x^j$, then the relation $P(x) = Cx^2 P''(x)$ implies that $a_j = Cj(j-1)a_j$ for all $j$;
hence $a_j = 0$ for $j \leq n-1$, and we conclude that $P(x) = a_n x^n$, which has all identical roots.