Solution
I received solutions from several of you. It is a simple exercise in complex numbers.
The idea is that if (a, b) is the final point of the path, then f(i) = bi + a, and this vanishes precisely when f(x) is divisible by (x − i) over the complex numbers, and therefore divisible by x² + 1 over the reals.
More detail: We may consider P(f) to be the sum of complex numbers, z_r, with r = 0...n. Noting that multiplication by i represents a counterclockwise rotation of 90°, we see that z_r = a(r) ir. The final point of the path is given by the sum of all z_r, which equals f(i). So the path lands at the origin precisely when f(i) = 0. Since f has real coefficients, this occurs precisely when f(−i) is also zero and so occurs precisely when f(x) is divisible by x² + 1.
The problem intrigued me because it is related to Lill's method of finding a root of polynomial. One makes the path P as in the statement of 1208. Then one launches another path, Q, but starting at an angle θ. This other path continues until it strikes the first path, at which point it turns 90° to the left and continues, always turning left when it strikes the first path. If θ is such that Q terminates at the terminus of P, then tanθ is a root of the initial polynomial.
Alberto Monteiro found the following extension: Consider the coefficient path in 3-space. Go a(0) units in the x-direction, then a(1) units in the y-direction, then a(2) units in the z-direction, etc. Then the path ends at the origin iff the polynomial is divisible by x³ + 1. He proved this by a method that is a little different than the method in the plane.
Fuller details of Lill's method can be found in the paper by Dan Kalman and Mark Verdi: Polynomials with closed Lill paths, Mathematics Magazine 88 (2015) 3-10.
[Back to Problem 1208]
© Copyright 2015 Stan Wagon. Reproduced with permission.