Diophantine equations are a certain class of equations for which solutions that consist only of integers are sought. So, for example, we know that Pythagoras’s theorem x^{2}+y^{2}=z^{2} has many sets of solutions such as the numbers x=3, y=4, z=5 or the set x=6, y=8, z=10. A lot of these problems involve existence claims such as whether any solution exists at all and if none can be found, whether it can be proved that no solution exists.

To prove that a solution exists, one can use brute force techniques to search through the space of integers and see if a set can be found that satisfies the equation. Nowadays the searches use computers. This was done recently in the case of the equation x^{3}+y^{3}+z^{3}=33, with the solution coming after 64 years of effort.

The solutions are.

x = +8,866,128,975,287,528

y = –8,778,405,442,862,239

z = –2,736,111,468,807,040

What is so special about this equation? As far as we know, nothing really, except that the numbers 33 and 42 were the only ones below 100 for which neither a solution nor a proof that no solution exists had been found. Such problems are viewed by mathematicians as challenges in themselves and they pursue them with great vigor. As the above link states, “When the news of his solution hit the internet earlier this month, fellow number theorists and math enthusiasts were feverish with excitement.” The appeal of solving a problem that has no immediate practical consequence can elude people who are not mathematicians. But it is akin to people who climb a mountain just because it is there, just because its existence poses a challenge.

This solution was found by a computer but in order to do so, Andrew Booker at the University of Bristol had to devise a special algorithm to target this particular equation with the number 33, rather than a more general approach.

So the only unsolved number that remains is 42. Whether this new algorithm can be modified to target that number is what people will try to see. But there is a more general result that is at stake. “A major result would be to prove the conjecture that k = x^{3}+y^{3}+z^{3} has infinitely many solutions for every whole number k, except those k that have a remainder of 4 or 5 after being divided by 9.”

The just-proved result for k=33 satisfies that conjecture since the remainder is 6. The remainder for 42 is also 6 so one would expect that a solution should exist.

consciousness razor says

z is negative

consciousness razor says

I could be misunderstanding something, but I think that’s still unknown. There’s no proof that it needs to be so, only a conjecture. Those were the types of numbers which remained because they’re the harder cases.

Of course, I would also be surprised if there’s no solution for k=42. I don’t know what kind of progress has been made for k>100.

Michael Sternberg says

The excellent YouTube channel Numberphile (which features pretty high-profile mathematicians) covered this problem in several episodes, the latest reporting on the recent finding:

https://www.numberphile.com/videos/42-is-the-new-33

Mano Singham says

consciousness razor @#1,

Thanks for catching it. I have corrected it.

consciousness razor says

Sorry, just to be clear….. by “types”, I meant ones like 33 and 42 and their ilk, which I don’t know how to specify other than to say they require large integers. Those are at least harder than ones that can easily be found with small integers. But it could be that, for reasons nobody understands yet, there’s no solution for 42.

However, it’s already known to be impossible, if the number is 4 or 5 (mod 9). So nobody should waste their time.

Lassi Hippeläinen says

It’s a pity that Douglas Adams isn’t available for comments.

Matt G says

Strange -- when I try to find the sum of three cubes I keep getting 3.

leftygomez says

The fact that numbers congruent to 4 or 5 (mod 9) can’t be written as the sum of 3 cubes is easy to see, since (by checking 0, 1, 2, … , 8) you can see that any integer cubed is congruent to one of -1, 0, or 1 (mod 9), and there is no way to get 4 or 5 (mod 9) by adding three chosen from -1, 0, and 1.

Mano Singham says

Thanks leftygomez,

That was very helpful.