Further correct solutions were submitted by Bradley alumnus Brian Laughlin (81') and from W.C.Soudriette, Augustin Murillo, Bill Webb, Ahron Teitelman, Ahmed abd Elmaksood, M. Lynch, Al Zimmermann, Paul Lee, Lou Cairoli, Paul Botham, Nancy Schwarzkopf, Nick McGrath, Ross Millikan, Burkart Venzke.

Let p(x) = a_n xⁿ+ a_{n -1} x^{n - 1}+ ... + a₂ x²+ a₁ x + a₀ be a palindromial whose derivative p'(x) = na_n x^{n - 1}+ (n - 1)a_{n - 1} x^{n - 2}+ ... + 2a₂ x + a₁  is also a palindromial.  From p(x) we get a_n= a₀ and from p'(x) we get na_n= a₁.  Now a_{n -1} is determined by p(x) and the value of a₁, from which in turn a₂ is determined from p'(x).  Continuing in this fashion, we see that all coefficients are determined by the value of a_n.  Since multiplying a palindromial by a constant doesn't change its citizenship as a palindromial, with the same true of its derivative, we conclude that there is a unique palindromial with leading coefficient 1 whose derivative is also a palindromial.  A moment's thought will verify that the polynomials (x + 1)ⁿ fit the requirements.

You are visitor number 2051 to this page.