Putnam 1999, Problem B-2

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.

Solution:
First solution: 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 $P(x) = x^kR(x)$ with $R(0) \neq 0$; a simple computation yields
$$P''(x) = (k^2-k)x^{k-2}R(x) + 2kx^{k-1}R'(x) + x^kR''(x).$$

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 $P(x) = \sum_{j=0}^n a_jx^j$, then the relation $P(x) = Cx^2P''(x)$ 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 $P(x) = a_nx^n$, which has all identical roots.

Second solution (by Greg Kuperberg): Let $f(x) = P''(x)/P(x) = 1/Q(x)$. By hypothesis, $f$ has at most two poles (counting multiplicity).

Recall that for any complex polynomial $P$, the roots of $P'$ lie within the convex hull of $P$. To show this, it suffices to show that if the roots of $P$ lie on one side of a line, say on the positive side of the imaginary axis, then $P'$ has no roots on the other side. That follows because if $r_1, \ldots, r_n$ are the roots of $P$,
$$\frac{P'(z)}{P(z)} = \sum_{i=1}^n \frac{1}{z-r_i}$$
and if $z$ has negative real part, so does $1/(z-r_i)$ for $i = 1, \ldots, n$, so the sum is nonzero.

The above argument also carries through if $z$ lies on the imaginary axis, provided that $z$ is not equal to a root of $P$. Thus we also have that no roots of $P'$ lie on the sides of the convex hull of $P$, unless they are also roots of $P$.

From this we conclude that if $r$ is a root of $P$ which is a vertex of the convex hull of the roots, and which is not also a root of $P'$, then $f$ has a single pole at $r$ (as $r$ cannot be a root of $P''$). On the other hand, if $r$ is a root of $P$ which is also a root of $P'$, it is a multiple root, and then $f$ has a double pole at $r$.

If $P$ has roots not all equal, the convex hull of its roots has at least two vertices.