Another Ridiculously Complicated Math Problem

Discus: SAT/ACT Tests and Test Preparation: July 2004 Archive: Another Ridiculously Complicated Math Problem
 By Apocalypse_Now (Apocalypse_Now) on Wednesday, July 14, 2004 - 02:49 pm: Edit

Each of the four equations

1^3 + 12^3 = 9^3 + 10^3

9^3 + 34^3 = 16^3 + 33^3

9^3 + 15^3 = 2^3 + 16^3

10^3 + 27^3 = 19^3 + 24^3

involves the cubes of four unequal positive integers. In each equation, the only positive integer that's a factor of all four integers is 1. There's a fifth such equation (whose terms are not merely a rearrangement of the terms of any of the above four equations) in which the sum of the cubes on each side is less than 64,000. What is this fifth equation?

 By Texas137 (Texas137) on Wednesday, July 14, 2004 - 03:11 pm: Edit

you might enjoy the math forums at www.artofproblemsolving.com

 By Al0 (Al0) on Wednesday, July 14, 2004 - 05:27 pm: Edit

2^3+34^3=15^3+33^3=39,312

2 != 34 != 15 != 33
The last two are odd and the first is 2, so they are relatively prime. This isn't just a re-arrangement of one of the above equations and the sum is 39,312 < 64,000

 By Apocalypse_Now (Apocalypse_Now) on Thursday, July 15, 2004 - 12:05 am: Edit

indeed