Problem A. 498. (January 2010)
A. 498. Let p(x) be a polynomial with integer coefficients, and let w be a complex number of unit absolute value. Prove that if the number c=p(w) is purely real, then there is a polynomial q(x) with integer coefficients for which .
Deadline expired on February 10, 2010.
Solution. If w=1 then p(w) is an integer, and the statement is trivial. So we can assume that w is not real.
Lemma. For every k nonnegative integer there exists a polynomial uk(x) such that
sin (k+1)t=uk(2cos t).sin t.
Proof. We apply induction on k.
In the cases k=0 and k=1 the polynomials u0(x)=1 and u1(x)=x satisfy the conditions, since sin t=1.sin t and sin 2t=2cos t.sin t.
If we already have uk-1(x) and uk-2(x) then define
This polynomial has integer coefficients too, and
sin (k+1)t=2sin ktcos t-sin (k-1)t=2.uk-1(2cos t)sin t.cos t-uk-2(2cos t)sin t=
=(uk-1(2cos t).2cos t-uk-2(2cos t))sin t=uk(2cos t)sin t.
The Lemma is complete
If then let .
If w=cos t+isin t, where by our assumption , then
Remark. The problem is closely related to the 6th problem of CIIM 1. In the solution above we used the ideas of Pablo Soberon Bravo (Mexiko).
7 students sent a solution. 5 points: Backhausz Tibor, Bodor Bertalan, Éles András, Nagy 235 János, Nagy 648 Donát, Strenner Péter, Szabó 928 Attila.