IMO 2002 SL N4

Origin: GER | Category: Number Theory

Problem

Is there a positive integer m such that the equation a + 1 b + 1 c + 1 abc = m a + b + c has infinitely many solutions in positive integers a, b, c?

Solution

For a = b = c = 1 we obtain m = 12. We claim that the given equation has infinitely many solutions in positive integers a, b, c for this value of m. After multiplication by abc(a+b+c) the equation 1 a+ 1 b+ 1 c+ 1 abc− a+b+c = 0 becomes a2(b + c) + b2(c + a) + c2(a + b) + a + b + c −9abc = 0. (1) We must show that this equation has infinitely many solutions in positive integers. Suppose that (a, b, c) is one such solution with a < b < c. Re- garding (1) as a quadratic equation in a, we see by Vieta’s formula that
b, c, bc+1 a  also satisfies (1).

Define (an)\infty n=0 by a0 = a1 = a2 = 1 and an+1 = anan−1+1 an−2 for each n > 1. We show that all an’s are integers. This procedure is fairly standard. The above relation for n and n −1 gives an+1an−2 = anan−1 + 1 and an−1an−2 + 1 = anan−3, so that adding yields an−2(an−1 + an+1) = an(an−1 + an−3). Therefore an+1+an−1 an = an−1+an−3 an−2 = \cdot \cdot \cdot , from which it follows that an+1 + an−1 an

. a2+a0 a1 = 2 for n odd; a3+a1 a2 = 3 for n even. It is now an immediate consequence that every an is integral. Also, the above consideration implies that (an−1, an, an+1) is a solution of (1) for each n \geq1. Since an is strictly increasing, this gives an infinity of solutions in integers. Remark. There are infinitely many values of m \inN for which the given equation has at least one solution in integers, and each of those values admits an infinity of solutions.