It is hard to seek answers in infinite space. High school math can help you narrow your search.
Given that people have been studying the properties of numbers for thousands of years, it might be decided that we know everything about number 3. However, recently mathematicians have discovered something new regarding number 3: a third way to express this number as the sum of three cubes. The task of writing a number through the sum of three cubes of integers is unexpectedly interesting. It is easy to show that most of the numbers cannot be written as a single cube or a sum of two cubes, but there is a hypothesis that most of the numbers can be written as a sum of three cubes. However, finding these cubes is sometimes extremely difficult.
For example, we knew that the number 3 can be written in the form 1 3 + 1 3 + 1 3 , and also in the form 4 3 + 4 3 + (-5) 3 , but for more than 60 years mathematicians have been interested in the question of whether there are more one way to do it. And this September, Andrew Booker and Andrew Sutherland finally found a third way :
3 = 569 936 821 221 962 380 720 3 + (−569 936 821 113 563 493 509) 3 + (−472 715 493 453 327 032) 3
If you want to check this result, do not try to use a calculator. Most of them will not cope with so many numbers. But WolframAlpha will handle this.
In search of new solutions for the number 3, mathematicians use the techniques invented this year by Booker, the first to find the sum of three cubes for the number 33. But why does such a break take so much time? In search of the right cubes, you have to cover a very large territory, and only a small number of tips can indicate the desired direction to us. Therefore, the trick is to find trickier search methods. To imagine the problem itself and its solution, we start with a simpler question: how can we write 33 as the sum of three integers?
We can write 33 = 19 + 6 + 8, or 33 = 11 + 11 + 11, or 33 = 31 + 1 + 1. We can also use negative numbers: 33 = 35 + (−1) + (−1). There are an infinite number of ways to do this, because you can always increase one or two numbers and decrease the third to compensate for this - for example, 33 = 36 + (−1) + (−2), 33 = 100 + 41 + (−108), and etc.
What about recording 33 as the sum of three squares? We will need to find numbers that are squares of integers, such as 1 = 1 2 , 9 = 3 2 , and 64 = 8 2 - their sum gives 33. Having played a little, you can find that 33 = 4 2 + 4 2 + 1 2 and 33 = 5 2 + 2 2 + 2 2 . Are there any other options? In principle, no. You can replace 4 with -4, and get 33 = (-4) 2 + 4 2 + 1 2 , which will give us some more ways to write our decisions, but no matter how you count them, there are not many ways to write 33 as the sum of three squares.
When summing squares, we do not have the same flexibility as when summing any integers. We have less choice, and, more importantly, addition only increases our amount. After all, the squares of integers are not negative - squaring both the positive and negative numbers always gives the positive.
Squares have more restrictions, but it also gives us something useful: our search space is "limited". Trying to find three squares giving a total of 33, we cannot use numbers whose squares are greater than 33, because as soon as our sum goes beyond 33, we will not be able to reduce it. And this means that we only need to consider combinations of 0 2 , 1 2 , 2 2 , 3 2 , 4 2 and 5 2 (their negative counterparts do not give us anything new, and we will ignore them).
Having six options for each of their three squares, we get no more than 6 × 6 × 6 = 216 ways to write 33 as their sum. A short list is enough to check all the possibilities and make sure that we have not missed anything.
Now back to the problem of the sum of three cubes. It is easy to see that it combines a limited choice from the problem of the sum of squares with an infinite search space from the problem of the sum of integers. As with squares, not every integer is a cube of another number. We can use numbers like 1 = 1 3 , 8 = 2 3 , 125 = 5 3 , but we cannot use 2, 3, 4, 10, 108, and most of the remaining numbers. But, unlike squares, cubes are negative - for example, (-2) 3 = -8, (-4) 3 = -64 - which means we can reduce our amount if necessary. Access to negative numbers gives us an unlimited number of options, that is, our search space, as in the case of the sum of integers, is unlimited.
The unlimited search space means that we can search for answers for a very long time. And people have been looking for them for decades. It took a supercomputer and tricky math to finally find the right combination of cubes. Let's see how this was done.
Suppose you need to find a solution to the equation:
33 = x 3 + y 3 + z 3
A simple approach is to mark out a certain region of numbers and substitute each of them until something comes up. If you find nothing, you can define a new search space and start over. This is similar to searching for new planets using a methodical study of the sky through a telescope.
Imagine that your initial search space limits all x, y and z to frames from -100 to 100. First you try:
(−100) 3 + (−100) 3 + (−100) 3
It didn’t work out. Then you try:
(−99) 3 + (−100) 3 + (−100) 3
Doesn't work either. You continue until you reach (100, −100, −100), then switch to (−100, −99, −100), and continue your hunt again. As a result, you will check about 200 × 200 × 200 = 8,000,000 options, not finding anything suitable. You have to designate a new search space and start over.
A more interesting approach is to rewrite the equation as follows:
33 - (x 3 + y 3 ) = z 3
Now, instead of iterating over all triples (x, y, z), we will iterating over deuces (x, y). For each pair, we will calculate the result, and then check the list of cubes, depending on whether our result is z 3 . If it is, a solution has been found. If not, we will continue to search. This greatly reduces the search space. Instead of 8,000,000 triples, we are now looking among 200 × 200 = 40,000 pairs. Serious savings, but still not enough to make the task computationally accessible.
An even more convenient approach is to rewrite the equation as follows:
33 - z 3 = x 3 + y 3
Now we iterate over z, and for each calculated z we use a tricky trick from a math course. The expression x 3 + y 3 can always be decomposed as follows:
x 3 + y 3 = (x + y) (x 2 - xy + y 2 )
This is the formula for the sum of the cubes. To test it, simply multiply the right side using the distribution rule:
(x + y) (x 2 - xy + y 2 ) = x 3 - x 2 y + xy 2 + yx 2 - xy 2 + y 3 = x 3 + y 3
How does this help us in our search? Counting 33 - z 3 , we decompose it into a product of primes, which computers can do well, at least in the number range of interest to us. Factoring 33 - z 3 into factors, we check whether it is possible to compose these factors in the form (x + y) (x 2 - xy + y 2 ). If so, we have found a solution.
Suppose, for example, that we are trying to write 34 as the sum of three cubes, and our searches led us to z = -6. We count 34 - z 3 = 34 - (-6) 3 = 34 - (-216) = 34 + 216 = 250, and now we decompose 250.
Having studied the question, we understand that we can write 250 = 10 × 25 = (5 + 5) (5 2 - 5 × 5 + 5 ² ). And this is exactly (x + y) (x 2 - xy + y 2 ) for x = 5 and y = 5, so the triple (x, y, z) = (5, 5, -6) should work for 34. And, of course, 34 = 5 3 + 5 3 + (-6) 3 , and we have successfully discovered three cubes, the sum of which gives 34.
Such a method allows, instead of 200 3 = 8,000,000 triples or even 200 2 = 40,000 pairs, to study 200 possible variants of z. The additional work is factorization and verification, but overall search efficiency is seriously increasing. Nevertheless, the search space studied in search of the sum of cubes giving a number such as 33 is so huge that even such improvements cannot help supercomputers get close to this task.
Then Andrew Booker stepped on the scene. He developed some additional techniques, using algebra and number theory, to further improve search efficiency. Having letting his university’s supercomputer do this, three weeks later he received the first found representation of 33 as the sum of three cubes:
33 = 8 866 128 975 287 528 3 + (−8 778 405 442 862 239) 3 + (−2 736 111 468 807 040) 3
Having solved this problem, before moving on to number 3, Booker and Sutherland solved the same problem for number 42:
42 = (−80 538 738 812 075 974) 3 + 80 435 758 145 817 515 3 + 12 602 123 297 335 631 3
You may be surprised that after thousands of years, we can still learn something new about numbers such as 3, 33 and 42. Perhaps even more surprising is that abstract things from school mathematics, such as the formula for sums, can help cubes. However, this is how mathematics works, and therefore we continue our research. So watch out for the number 114 - the smallest of the numbers for today, for which the sum of three cubes has not yet been found. I have a feeling that the search has already begun for Andrew Booker and other mathematicians.