The cube of an integer is congruent modulo 9 to either 0, 1 or −1. To see this, simply check the integers from 1 to 9. Modulo 9, the sum of three cubes can therefore take any value from 0 to 8 except 4 and 5.
Conversely, given an integer k whose remainder upon division by 9 is neither 4 nor 5, can it be written as the sum of three cubes of integers? In other words, can we find three integers a, b and c such that k = a3 + b3 + c3? The answer is believed to be yes. For some values of k, this is easy, and it can sometimes be done in several ways. For example, 3 = 13 + 13 + 13 = 43 + 43 + (−5)3.
This conjecture, proposed in 1955 by the British mathematician Louis Joel Mordell (1888–1972), has still not been proved. But for values of k from 0 to 100, only two stubborn holdouts remained: 33 and 42.
Earlier this year, using sophisticated algorithms and immensely powerful computers, Andrew Booker of the University of Bristol in England showed that:
33 = (8,866,128,975,287,528)3
+ (−877,845,442,862,239)3
+ (−27,361,114,688,070,040)3.
Very recently, Andrew Sutherland, an American mathematician at MIT in Boston, Massachusetts, discovered that:
42 = (−80,538,738,812,075,974)3
+ 80,435,758,145,817,5153
+ 12,602,123,297,335,6313.
The conjecture has therefore been proved for every integer below 100. However, there are still eleven numbers below 1,000 for which no solution is known: 114, 165, 390, 579, 627, 633, 732, 795, 906, 921 and 975.
