IMO 1976 SL 9

Origin: FIN

Problem

Let P1(x) = x2 −2, Pj(x) = P1(Pj−1(x)), j = 2, 3, . . . . Show that for arbitrary n the roots of the equation Pn(x) = x are real and different from one another.

Solution

The equation Pn(x) = x is of degree 2n, and has at most 2n distinct roots. If x > 2, then by simple induction Pn(x) > x for all n. Similarly, if x < −1, then P1(x) > 2, which implies Pn(x) > 2 for all n. It follows that all real roots of the equation Pn(x) = x lie in the interval [−2, 2], and thus have the form x = 2 cost. Now we observe that P1(2 cos t) = 4 cos2 t −2 = 2 cos 2t, and in general Pn(2 cost) = 2 cos 2nt. Our equation becomes cos 2nt = cos t, which indeed has 2n different solutions t = 2\pim 2n−1 (m = 0, 1, . . . , 2n−1 −1) and t = 2\pim 2n+1 (m = 1, 2, . . ., 2n−1).