### 31st Putnam 1970

Problem A2

p(x, y) = a x2 + b x y + c y2 is a homogeneous real polynomial of degree 2 such that b2 < 4ac, and q(x, y) is a homogeneous real polynomial of degree 3. Show that we can find k > 0 such that p(x, y) = q(x, y) has no roots in the disk x2 + y2 < k except (0, 0).

Solution

The disk is a strong hint that one should use polar coordinates, so put x = r cos θ, y = r sin θ. Then we get r = p(cos θ, sin θ)/q(cos θ, sin θ). Now |cos θ| and |sin θ| ≤ 1, so |q(cos θ, sin θ)| ≤ h, where h is the sum of the absolute values of the coefficients of q. So we are home if we can establish some inequality |p(cos θ, sin θ)| ≥ h', because we can then take k = h'/h.

Without loss of generality a > 0 and hence also c > 0. Now p(cos θ, sin θ) = 1/2 (a + c)(cos2θ + sin2θ) + 1/2 (a - c)(cos2θ - sin2θ) + 1/2 b sin 2θ = 1/2 (a + c) + 1/2 (a - c) cos 2θ + 1/2 b sin 2θ. Put d = √(b2 + (a - c)2) and take φ so that cos φ = (a - c)/d, sin φ = b/d. Then 2p(cos θ, sin θ) = (a + c) + d (cos φ cos 2θ + sin φ sin 2θ) = (a + c) + d cos(φ - 2θ) >= (a + c) - d.

But 4ac > b2, so (a + c)2 > (a - c)2 + b2, so (a + c) > d, and we are done.