The number 145 is a strong number

Douglas Adams: The Secrets of the Number 42

83 = 8 (mod 9)

That means, cube numbers modulo 9 are either 8, 0, or 1. The addition of three cube numbers in the restricted number space results in:

0 = 0 + 0 + 0 = 0 + 1 + 8 

1 = 1 + 0 + 0 = 1 + 1 + 8 

2 = 1 + 1 + 0 

3 = 1 + 1 + 1 

6 = 8 + 8 + 8

7 = 8 + 8 + 0 

8 = 8 + 0 + 0 = 1 + 8 + 8

You can never get a 4 or a 5 with the cube numbers. This means that the sums of three cubic numbers never have the form 9m + 4 or 9m + 5, which is why they are called "forbidden values".

Search for solutions

The search for solutions to the equation n = a3 + b3 + c3 is extremely complicated. This is already shown by the cases n = 1 and n = 2:

For n = 1 there is the obvious solution 13 + 13 + (-1)3 = 1. But there are also others:

93 + (-6)3 + (-8)3 = 729 + (-216) + (-512) = 1

And it goes further. In 1936 the German mathematician Kurt Mahler identified an infinite number of solutions:

(9p4)3 + (3p - 9p4)3 + (1 - 9p3)3 = 1, where p is any integer.

N = 2 also has an infinite number of solutions. The mathematician A.S. Werebrusow discovered them as early as 1908:

(6p3 + 1)3 + (1 - 6p3)3 + (-6p2)3 = 2.

By taking the terms of both equations with a cube number r3 multiplied, one gets for every cube number r3 and every double cube number 2r3 infinite solutions. For example, for 16, twice the cube of 2, with p = 1:

143 + (-10)3 + (-12)3 = 16

Until August 2019, however, only two solutions were known for n = 3: 13 + 13 + 13 = 3 and 43 + 43 + (-5)3 = 3. But with enormous effort, a third possibility was finally found:

3 = (-472 715 493 453 327 032)3 + (-569 936 821 113 563 493 509)3 + 569 936 821 221 962 380 7203

But what about other values ​​of n that are not among the forbidden numbers? Do you always have a solution?

Hardworking computers

To answer that, one can examine the possible values ​​individually: 1, 2, 3, 6, 7, 8, 9, 10, 11, 12, 15, 16, ... (this sequence is also noted in Sloane's encyclopedia as A060464 ). If the project succeeds in all cases, one would have an indication that every whole number n that does not have the form n = 9m + 4 or n = 9m + 5 can be broken down into the sum of three cube numbers.

In fact, mathematicians have now achieved impressive results with the help of powerful computer networks. But some puzzles remained unsolved - and that leads us once again to the fascinating number 42.

In 2009, the mathematicians Andrea-Stephan Elsenhans and Jörg Jahnel examined all triplets a, b, c with amounts less than 1014to find solutions for n between 1 and 1000. Only 14 numbers had no solution in this area: 33, 42, 74, 114, 165, 390, 579, 627, 633, 732, 795, 906, 921 and 975 no solution. For n less than 100, only three numbers remained open: 33, 42 and 74.

In 2016 Sander Huismen identified a solution for 74: (-284 650 292 555 885)3 + (66 229 832 190 556)3 + (283 450 105 697 727)3. And in March 2019, Andrew Booker settled Case 33: (8 866 128 975 287 528)3 + (-877 8405 442 862 239)3 + (-2 736 111 468 807 040)3.

Suddenly 42 was the last positive whole number less than 100 that you didn't know whether it could be written as the sum of three cube numbers. Should that not work, there would have to be a compelling mathematical reason for it - that would make 42 extremely unusual. The computers were running at full speed, but the problem seemed overwhelming.

The answer finally came in September 2019 - just six months after the solution for the 33rd was found. It was the result of a huge calculation coordinated by Andrew Booker and Andrew Sutherland. The enormous achievement was achieved by several private computers that were part of the "Charity Engine" network, which makes the computing time available to universities and other projects. After a total of more than a million hours of computing time, they found the following result:

42 = (– 80 538 738 812 075 974)3 + 80 435 758 145 817 5153 + 12 602 123 297 335 6313

The cases for n = 165, 795, and 906 were also recently resolved. This leaves only 114, 390, 579, 627, 633, 732, 921 and 975 open for numbers below 1000. The suspicion that there are solutions for all n that are not in the form 9m + 4 or 9m + 5 is becoming more and more solid. In 1992 Roger Heath-Brown proposed a stronger conjecture, according to which there would be an infinite number of possibilities for all non-prohibited n to write them as the sum of three cube numbers. So the work is far from over.

Because the problem is so difficult, some mathematicians assume that it can be undecidable. Then no algorithm, however clever, would be able to handle all possible cases of n. A typical example of an undecidable problem is the so-called hold problem, which Alan Turing presented in 1936: There is no algorithm that can decide for any given program whether its computation ends at some point or whether it continues forever. But the question about the sum of cubic numbers is different: This is a purely mathematical question that is easy to explain. If one were to prove such undecidability for this, that would be something completely new.

The sum of three cube numbers is now also solved for 42, but it was certainly not the last problem in which the famous number appears.

Note: In an earlier version of the text it was stated that the 42 kilometers of the marathon had its roots in ancient Greece. That's not true. We have therefore removed the notice.