IMO 1989 SL 9

Origin: FRA

Problem

For all integers n, n \geq0, there exist uniquely determined integers an, bn, cn such that  1 + 4 3\sqrt 2 −4 3\sqrt n = an + bn 3\sqrt 2 + cn 3\sqrt 4. Prove that cn = 0 implies n = 0.

Solution

From an+1 + bn+1 3\sqrt 2 + cn+1 3\sqrt 4 = (an + bn 3\sqrt 2 + cn 3\sqrt 4)(1 + 4 3\sqrt 2 −4 3\sqrt 4) we obtain an+1 = an −8bn + 8cn. Since a0 = 1, an is odd for all n. For an integer k > 0, we can write k = 2lk′, k′ being odd and l a nonneg- ative integer. Let us set v(k) = l, and define \betan = v(bn), \gamman = v(cn). We prove the following lemmas: Lemma 1. For every integer p \geq0, b2p and c2p are nonzero, and \beta2p = \gamma2p = p + 2. Proof. By induction on p. For p = 0, b1 = 4 and c1 = −4, so the assertion is true. Suppose that it holds for p. Then (1+4 3\sqrt 2−4 3\sqrt 4)2p+1 = (a+2p+2(b′ 3\sqrt 2+c′ 3\sqrt 4))2 with a, b′, and c′ odd. Then we easily obtain that (1 + 4 3\sqrt 2 −4 3\sqrt 4)2p+1 = A + 2p+3(B 3\sqrt 2 + C 3\sqrt 4), where A, B = ab′ +2p+1E, C = ac′ +2p+1F are odd. Therefore Lemma 1 holds for p + 1. Lemma 2. Suppose that for integers n, m \geq0, \betan = \gamman = \lambda > \betam = \gammam = µ. Then bn+m, cn+m are nonzero and \betan+m = \gamman+m = µ. Proof. Calculating (a′ + 2\lambda(b′ 3\sqrt 2 + c′ 3\sqrt 4))(a′′ + 2µ(b′′ 3\sqrt 2 + c′′ 3\sqrt 4)), with a′, b′, c′, a′′, b′′, c′′ odd, we easily obtain the product A + 2µ(B 3\sqrt 2 + C 3\sqrt 4), where A, B = a′b′′ + 2\lambda−µE, and C = a′c′′ + 2\lambda−µF are odd, which proves Lemma 2. Since every integer n > 0 can be written as n = 2pr + \cdot \cdot \cdot + 2p1, with 0 \leqp1 < \cdot \cdot \cdot < pr, from Lemmas 1 and 2 it follows that cn is nonzero, and that \gamman = p1 + 2. Remark. b1989 and c1989 are divisible by 4, but not by 8.