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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 c=q\left(w+\frac1w\right).

(5 pont)

Deadline expired on 10 February 2010.


Solution. If w=\pm1 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

uk(x)=uk-1(x).x-uk-2(x).

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 \displaystyle p(x)=\sum_{k=0}^na_kx^n then let \displaystyle q(x) = \sum_{k=0}^n a_k u_k(x).

If w=cos t+isin t, where by our assumption \sin
t=\mathrm{Im}\;w\ne0, then


  q\left(w+\frac1w\right) = q(2\cos t) = 
  \sum_{k=0}^n a_k u_k(2\cos t) =
  \sum_{k=0}^n a_k \frac{\sin(k+1)t}{\sin t} = 
  \sum_{k=0}^n a_k \frac{\cos kt \sin t + \sin kt \cos t}{\sin t} =


  = \sum_{k=0}^n a_k \left(\cos kt + \frac{\cos t}{\sin t} \sin kt \right)
  = \sum_{k=0}^n a_k \cos kt + \frac{\cos t}{\sin t} \sum_{k=0}^n a_k  \sin kt
  = \mathrm{Re}\; p(w) + \frac{\cos t}{\sin t} \mathrm{Im}\; p(w)
  = p(w).

 

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).


Statistics:

7 students sent a solution.
5 points:Backhausz Tibor, Bodor Bertalan, Éles András, Nagy 235 János, Nagy 648 Donát, Strenner Péter, Szabó 928 Attila.

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