14th Irish 2001 Problem B4 Find all non-negative reals for which (13 + √x) 1/3 + (13 - √x) 1/3 is an integer.
x = 137 53/216, 168 728/729, 196, 747 19/27 Solution Put f(x) = (13 + √x) 1/3 + (13 - √x) 1/3 . The first step is to show that f(x) is a strictly decreasing function of x. Suppose A 3 = (√x + 13), (A-k) 3 = (√x - 13), and B 3 = (√y + 13). Suppose further that x  0. It may be that A - k  0. Hence 3k(B-A)(B+A-k) > 0, so 3kB 2 - 3k 2 B > 3kA 2 - 3k 2 A or B 3 - (B-k) 3 > A 3 - (A-k) 3 = 26. Hence B-k  0. We have f(0) = 2·13 1/3 , and so 4  -13 1/3 > -2.5, so h = -2 + (√30)/6. Hence h 3 + 13 = (77/36)√30, so x = 137 53/216. Similarly, if (√x - 13) 1/3 = h, (√x + 13) 1/3 = h+3. Then (h+3) 3 = h 3 + 26, so 9h 2 + 27h + 1 = 0, so h = (-27 + (√693) )/18 = (-9 + √77)/6. Hence h 3 + 13 = (40/27)√77, so x = 168 728/729. Similarly, if (√x - 13) 1/3 = h, (√x + 13) 1/3 = h+2. Then (h+2) 3 = h 3 + 26, so h 2 + 2h - 3 = 0 and hence h = 1. So x = 196. Finally, if (√x - 13) 1/3 = h, (√x + 13) 1/3 = h+1. Then (h+1) 3 = h 3 + 26, so 3h 2 + 3h - 25 = 0 and hence h = (-3 + √309)/6, giving x = 747 19/27. 14th Irish 2001 © John Scholes jscholes@kalva.demon.co.uk 16 December 2003 Last updated/corrected 16 Dec 03