7th Brazil 1985 Problem 5 A, B are reals. Find a necessary and sufficient condition for Ax + B[x] = Ay + B[y] to have no solutions except x = y.
A = 0 or B/A ∈ (-2, 0) Solution If A = 0, then we have B[x] = B[y] which obviously has infinitely many solutions with x ≠ y. So assume A ≠ 0. Then we can write the equation as B/A ([x] - [y]) = y - x. If x ≠ y, we can assume x  0, so [x] < [y]. Write x = n + ε, y = n + d + δ, where n is any integer, d is a non-negative integer and δ, ε ∈ [0, 1). So B/A = - (d + η)/(d+1), where d is a non-negative integer and η = δ+ε ∈ [0, 2). We note first that -2 < -(d + η)/(d+1) ≤ 0. But in fact we cannot get the value 0 without η = d = 0 and then x = y. So if there are solutions with x ≠ y and A ≠ 0, then B/A must belong to the open interval (-2, 0). But conversely by taking d = 0 and suitable η it is clear that B/A can take any value in (-2,0). Thus a necessary and sufficient condition for solutions with x ≠ y is A = 0 or B/A ∈ (-2, 0). 7th Brazil 1985 © John Scholes jscholes@kalva.demon.co.uk 24 Oct 2003 Last corrected/updated 24 Oct 03