IMO 1973 SL 16

Origin: CUB

Problem

Given a, \theta \inR, m \inN, and P(x) = x2m −2|a|mxm cos \theta+a2m, factorize P(x) as a product of m real quadratic polynomials.

Solution

First, we have P(x) = Q(x)R(x) for Q(x) = xm −|a|mei\theta and R(x) = xm −|a|me−i\theta, where eiϕ means of course cos ϕ + i sin ϕ. It remains to factor both Q and R. Suppose that Q(x) = (x −q1) \cdot \cdot \cdot (x −qm) and R(x) = (x −r1) \cdot \cdot \cdot (x −rm). Considering Q(x), we see that |qm k | = |a|m and also |qk| = |a| for k = 1, . . . , m. Thus we may put qk = |a|ei\betak and obtain by de Moivre’s formula qm k = |a|meim\betak. It follows that m\betak = \theta + 2j\pi for some j \inZ, and we have exactly m possibilities for \betak modulo 2\pi: \betak = \theta+2(k−1)\pi m for k = 1, 2, . . ., m. Thus qk = |a|ei\betak; analogously we obtain for R(x) that rk = |a|e−i\betak. Consequently, xm−|a|mei\theta = m

k=1 (x−|a|ei\betak) and xm−|a|me−i\theta = m

k=1 (x−|a|e−i\betak). Finally, grouping the kth factors of both polynomials, we get P(x) = m

k=1 (x −|a|ei\betak)(x −|a|e−i\betak) = m

k=1 (x2 −2|a|x cos \betak + a2)

m

k=1  x2 −2|a|x cos \theta + 2(k −1)\pi m

• a2  .