42 is the new 33

A most interesting article appeared in Quanta magazine recently titled: Sum-of-Three-Cubes Problem Solved for ‘Stubborn’ Number 33. The article begins:

Mathematicians long wondered whether it’s possible to express the number 33 as the sum of three cubes — that is, whether the equation 33 = x³+ y³+ z³ has a solution. They knew that 29 could be written as 3³ + 1³ + 1³, for instance, whereas 32 is not expressible as the sum of three integers each raised to the third power. But the case of 33 went unsolved for 64 years. Now, Andrew Booker, a mathematician at the University of Bristol, has finally cracked it: 

He discovered that 

(8,866,128,975,287,528)³ + (–8,778,405,442,862,239)³ + (–2,736,111,468,807,040)³ = 33.

Apparently he used a very efficient search algorithm but he still required the use of a supercomputer running for three weeks to come up with the solution. Reading the article I also learned that there are no integer solutions to the equation x³+ y³+ z³ = n if \(n\equiv 4 \) mod 9 or \(n \equiv 5\) mod 9. Thus 31 and 32 cannot be expressed as the sums of three cubes. Of course, I've no idea why this is so and I may investigate the reason at some point in the future.

Interestingly, the only number below 100, for which a representation as a sum of three cubes has not been found, is 42. This is a number that featured in my blog post about magic cubes. Between 101 and 1000, there are 11 other "stubborn" numbers for which a representation has not been found. These numbers are 114, 165, 390, 579, 627, 633, 732, 795, 906, 921, 975. All of these numbers have the property in common that \(n\equiv 3 \) mod 9 or \(n \equiv 6\) mod 9.

The equation x³+ y³+ z³ = n is an example of a Diophantine equation (a polynomial equation whose unknown variables must take integer values). For mathematicians, "A major result would be to prove the conjecture that n = x³ + y³ + z³ has infinitely many solutions for every whole number \(n\), except those \(n\) that have a remainder of 4 or 5 after being divided by 9."

Figure 1: solutions for values of m
between 1 and 10

My source for the following information comes from here. For some such Diophantine equations, there are infinitely many solutions. For example, x³+ y³+ z³ = 1 has infinitely many solutions because of the identities:$$(1 + 9m^3)^3 + (9m^4)^3 + (-9m^4 - 3m)^3 = 1$$ $$(1 - 9m^3)^3 + (9m^4)^3 + (-9m^4 + 3m)^3 = 1$$By assigning various integer values to \(m\), the solutions unfold. Figure 1 shows an example using values of m between 1 and 10.

If we write the equation in the form:$$x^3 + y^3 + z^3 = t^3$$ and set \(t\) to be an integer then Ramanujan found that:$$x = 3n^2 + 5nm - 5m^2$$ $$y = 4n^2 - 4nm + 6m^2$$ $$z = 5n^2 - 5nm - 3m^2$$ $$t = 6n^2 - 4nm + 4m^2$$If we set \(m=1\) and \(n=2\), then \(t=20\) and \(t^3=8000\). Thus for the equation:$$x^3+y^3+z^3=8000$$there is the solution \(x=17\), \(y=14\) and \(z=7\) or $$17^3+14^3+7^3=8000$$While these results are interesting, they clearly only work for certain numbers and provide no help at all in solving a still outstanding problem like:$$x^3+y^3+z^3=42$$A specific curiosity that should be mentioned is:$$ 3^3 + 4^3 + 5^3 = 6^3$$Another is the smallest cube number that is the sum of three different positive cubes. This turns out to be 216 or 6^3 and itself the sum of \( -3^3+6^3+3^3 \). 729 or \( 9^3 \) is the next such number since \(8^3 + 6^3 + 1^1 = 729\). 729 is a perfect cube and a perfect square since \(27^2=729\).

Here is the excellent Numberphile video on YouTube in which Andrew Booker talks about his discovery:

UPDATE (7th September 2019): a solution for 42 has been found: