IMO 2012 Shortlist N4

Category: Number Theory

Problem

An integer a is called friendly if the equation (m2

n)(n2
m) = a(m − n)3 has a solution over the positive integers. a) Prove that there are at least 500 friendly integers in the set {1,2,...,2012}. b) Decide whether a = 2 is friendly. Solution. a) Every a of the form a = 4k − 3 with k ≥ 2 is friendly. Indeed the numbers m = 2k − 1 > 0 and n = k − 1 > 0 satisfy the given equation with a = 4k − 3: (m2
n)(n2
m) = (2k − 1)2
(k − 1) (k − 1)2
(2k − 1) = (4k − 3)k3 = a(m − n)3 . Hence 5,9,...,2009 are friendly and so {1,2,...,2012} contains at least 502 friendly numbers. b) We show that a = 2 is not friendly. Consider the equation with a = 2 and rewrite its left-hand side as a difference of squares: (m2
n + n2
m)2 − (m2
n − n2 − m)2 = 2(m − n)3 . Since m2
n − n2 − m = (m − n)(m + n − 1), we can further reformulate the equation as (m2
n + n2
m)2 = (m − n)2 8(m − n) + (m + n − 1)2 . It follows that 8(m − n) + (m + n − 1)2 is a perfect square. Clearly m > n, hence there is an integer s ≥ 1 such that (m + n − 1 + 2s)2 = 8(m − n) + (m + n − 1)2 . Subtracting the squares gives s(m + n − 1 + s) = 2(m − n). Since m + n − 1 + s > m − n, we conclude that s < 2. Therefore the only possibility is s = 1 and m = 3n. However then the left-hand side of the given equation (with a = 2) is greater than m3 = 27n3 , whereas its right-hand side equals 16n3 . The contradiction proves that a = 2 is not friendly. Comment. A computer search shows that there are 561 friendly numbers in {1,2,...,2012}.46