33rd Putnam 1972

Problem B6

The polynomial p(x) has all coefficients 0 or 1, and p(0) = 1. Show that if the complex number z is a root, then |z| ≥ (√5 - 1)/2.

Solution

Let h = (√5 - 1)/2. h is approx 0. 618, so certainly if |z| ≥ 1, then |z| > h. So it is sufficient to show that if z is a root with |z| < 1, then |z| ≥ h.

For such values of z, 1 + z + z2 + ... converges, so we have 2 + z + z2 + z3 + ... - 2 p(z) = ± z ± z2 ± z3 ± ... . Now |rhs| ≤ |z| + |z2| + ... = |z|/(1 - |z| ). If z is a root, then p(z) = 0, so lhs = 2 + z + z2 + ... = 2 + z/(1 - z) . If we could show that |lhs| ≥ (2 + |z|)/(1 + |z|), then we would have |z|/(1 - |z|) ≥ (2 + |z|)/(1 + |z|) and hence |z|2 + |z| - 1 ≥ 0. But |z| ≥ 0, so |z| ≥ h.

We require |2 - z| (1 + |z|) ≥ |1 - z| (2 + |z|). Squaring and using the polar form z = r e, this is equivalent to (4 - 4r cos θ + r2)(1 + r)2 ≥ (1 - 2r cos θ + r2) (2 + r)2, or after some simplification, 2r (1 - r2)(1 + cos θ) ≥ 0, which is true.

Many thanks to Thomas Linhart for correcting an error in the original proof.