Prove that if n is a positive integer such that the equation
x3 - 3xy2 + y3 = n
has a solution in integers x, y, then it has at least three such solutions. Show that the equation has no solutions in integers for n = 2891.
Solution
If x, y is a solution then so is y-x, -x. Hence also -y, x-y. If the first two are the same, then y = -x, and x = y-x = -2x, so x = y = 0, which is impossible, since n > 0. Similarly, if any other pair are the same.
2891 = 2 (mod 9) and there is no solution to x3 - 3xy2 + y3 = 2 (mod 9). The two cubes are each -1, 0 or 1, and the other term is 0, 3 or 6, so the only solution is to have the cubes congruent to 1 and -1 and the other term congruent to 0. But the other term cannot be congruent to 0, unless one of x, y is a multiple of 3, in which case its cube is congruent to 0, not 1 or -1.
Solutions are also available in Murray S Klamkin, International Mathematical Olympiads 1978-1985, MAA 1986, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
14 Oct 1998