From outside Bradley University, correct solutions were submitted by Monty Gray, Bill Webb, Lorenzo Pozzoli, Emanuele Macri, Massimo Brignone, Burkart Venzke, Tim Kelley, Philippe Fondanaiche.

The first polynomial is p₀ (x) = 1, which clearly has no root.  We proceed by induction, assuming that p_2n(x) has no root and showing that p_2n+2(x) has no root.  The above two students both observed that p_2n+2(x)'' = p_2n(x), so p_2n+2(x) is concave upward throughout and so has at most one minimum value which must occur at the point a where p_2n+2(a)' = 0.  But (aha!) p_2n+2(x)' = p_2n+1(x) and  p_2n+2(x)  = p_2n+1(x)+x²ⁿ⁺²/(2n+2)!, so p_2n+2(a) = a²ⁿ⁺²/(2n+2)! > 0.  Hence the minimum value is positive and p_2n+2(x) is root-free.

Several solvers invoked Taylor's remainder theorem for the exponential function e^x, and one correspodent threatened to use the incomplete Gamma function.

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ã2000 Alberto L. Delgado