Solution: Day 3, problem 2
We will say that a polynomial with real coefficients splits if all its roots are real.
The first observation is that, if h(x) splits, then so does (d/dx - a)h(x) for any a in R.
To see this, suppose that a is non-zero (the case a = 0 can be
done by a similar argument, or by continuity), and consider the polynomial h1(x) = h(x) - a-1 h'(x).
If h has degree n and n distinct roots x1 < ... < xn, then h has alternating signs in the
n + 1 intervals into which these roots divide R, and has a maximum or minimum in each of these intervals
(at infinity for the two end intervals). But h1 has the same asymptotics as h for x going to infinity
(because it has the same degree and the same leading coefficient as h), and has the same value and hence the same
sign as h at all maxima or minima of h, so it also has n sign changes and therefore splits. If there are
repeated roots, we modify the argument slightly (if h has an r - fold root at xi with r > 1, then h1
has an (r - 1) - fold root there and at least one further sign change between xi and the nearest maximum or
minimum of h on the left or right, because r - 1 has the opposite parity from r, so we still get as many
real roots as we need) or argue by continuity.
The next step is: if g(x) and h(x) both split, then so does
g(d/dx)(h(x)). Indeed, g(x) can be written as c (x - a1) ... (x - ar) with some non-zero real number c and real ai,
so the result follows from the previous one by induction on r. Now assume that bnxn + bn-1xn-1 + ... + b0 splits and apply the assertion just proved to g(x) = bnxN-n + bn-1xN-n+1 + ... + b0xN and h(x) = xN/N!, which both split.
Return to the problems.
Don Zagier 1996