43rd Swedish 2003 Problem 3 Which reals x satisfy [x 2 - 2x] + 2[x] = [x] 2 ?
all negative integers, 0, all intervals [1,1+√2), [3, √5+1), ... , [n,√(n 2 -2n+2)+1), ... Solution Consider first x ≥ 1. Note that n-1  [x] 2 . If 0 < x < 1, then -1 < x 2 - 2x < 0, so [x 2 -2x] = -1, but [x] = 0, so [x 2 - 2x] + 2[x] = -1 < 0 = [x] 2 . If x = 0, then obviously [x 2 - 2x] + 2[x] = [x] 2 . It remains to consider x < 0. Suppose -n < x < -n+1. Then [x] = -n. But x 2 - 2x < n 2 + 2n, so [x 2 - 2x] < n2 + 2n. Hence [x 2 - 2x] + 2[x] < n 2 = [x] 2 . So for x < 0 the only solutions are x an integer. 43rd Swedish 2003 © John Scholes jscholes@kalva.demon.co.uk 13 January 2004 Last corrected/updated 13 Jan 04