IMO 1970 SL 11

Origin: SWE

Problem

Let P, Q, R be polynomials and let S(x) = P(x3) + xQ(x3) + x2R(x3) be a polynomial of degree n whose roots x1, . . . , xn are distinct. Construct with the aid of the polynomials P, Q, R a polynomial T of degree n that has the roots x3 1, x3 2, . . . , x3 n.

Solution

Let S(x) = (x−x1)(x−x2) \cdot \cdot \cdot (x−xn). We have x3 −x3 i = (x−xi)(\omegax− xi)(\omega2x −xi), where \omega is a primitive third root of 1. Multiplying these equalities for i = 1, . . . , n we obtain T (x3) = (x3 −x3 1)(x3 −x3 2) \cdot \cdot \cdot (x3 −x3 n) = S(x)S(\omegax)S(\omega2x). Since S(\omegax) = P(x3) + \omegaxQ(x3) + \omega2x2R(x3) and S(\omega2x) = P(x3) + \omega2xQ(x3) + \omegax2R(x3), the above expression reduces to T (x3) = P 3(x3) + x3Q3(x3) + x6R3(x3) −3P(x3)Q(x3)R(x3). Therefore the zeros of the polynomial T (x) = P 3(x) + xQ3(x) + x2R3(x) −3P(x)Q(x)R(x) are exactly x3 1, . . . , x3 n. It is easily verified that deg T = deg S = n, and hence T is the desired polynomial.