IMO 1971 SL 1

Origin: BUL

Problem

Consider a sequence of polynomials P0(x), P1(x), P2(x), . . . , Pn(x), . . . , where P0(x) = 2, P1(x) = x and for every n \geq1 the following equality holds: Pn+1(x) + Pn−1(x) = xPn(x). Prove that there exist three real numbers a, b, c such that for all n \geq1, (x2 −4)[P 2 n(x) −4] = [aPn+1(x) + bPn(x) + cPn−1(x)]2. (1)

Solution

Assuming that a, b, c in (1) exist, let us find what their values should be. Since P2(x) = x2 −2, equation (1) for n = 1 becomes (x2 −4)2 = [a(x2 −2) + bx + 2c]2. Therefore, there are two possibilities for (a, b, c): (1, 0, −1) and (−1, 0, 1). In both cases we must prove that (x2 −4)[Pn(x)2 −4] = [Pn+1(x) −Pn−1(x)]2. (2) It suffices to prove (2) for all x in the interval [−2, 2]. In this interval we can set x = 2 cost for some real t. We prove by induction that Pn(x) = 2 cosnt for all n. (3) This is trivial for n = 0, 1. Assume (3) holds for some n −1 and n. Then Pn+1(x) = 4 cost cos nt −2 cos(n −1)t = 2 cos(n + 1)t by the additive formula for the cosine. This completes the induction. Now (2) reduces to the obviously correct equality 16 sin2 t sin2 nt = (2 cos(n + 1)t −2 cos(n −1)t)2. Second solution. If x is fixed, the linear recurrence relation Pn+1(x) + Pn−1(x) = xPn(x) can be solved in the standard way. The characteristic polynomial t2 −xt + 1 has zeros t1,2 with t1 + t2 = x and t1t2 = 1; hence, the general Pn(x) has the form atn 1 + btn 2 for some constants a, b. From P0 = 2 and P1 = x we obtain that Pn(x) = tn 1 + tn 2. Plugging in these values and using t1t2 = 1 one easily verifies (2).