Prove that the set of points satisfying x4 - x2 = y4 - y2 = z4 - z2 is the union of 4 straight lines and 6 ellipses.
x4 - x2 = (x2 - 1/2)2 - 1/4, so x4 - x2 = y4 - y2 is equivalent to (x2 - 1/2)2 = (y2 - 1/2)2 and hence to x2 - 1/2 = +/- (y2 - 1/2), which is equivalent to x = y, or x = -y, or x2 + y2 = 1. Similarly, for y4 - y2 = z4 - z2. So the equation given is equivalent to: (1) x = y and y = z, or (2) x = y and y = -z, or (3) x = -y and y = z, or (4) x = -y and y = -z, or (5) x = y and y2 + z2 = 1, or (6) x = -y and y2 + z2 = 1, or (7) x2 + y2 = 1 and y = z, or (8) x2 + y2 = 1 and y = -z, or (9) x2 + y2 = 1 and y2 + z2 = 1.
Clearly (1) - (4) are straight lines. (5) is the intersection of a plane and a cylinder, which is an ellipse. Similarly (6), (7) and (8). (9) is slightly harder to see. If one's visualization is good, then one can see that the intersection of two cylinders with the same radius and axes intersecting at right angles is two perpendicular ellipses with a common minor axis. Otherwise, subtracting the two equations we see that x = z or -z and the intersection is also given by the intersection of a cylinder with two planes.
3rd Putnam 1940
© John Scholes
15 Sep 1999