## 33 is the sum of three cubes

I’m a bit late to the party, but I find this fascinating: we now know (thanks to a discovery of Andrew R. Booker) that the number 33 can be written as the sum of three cubes. This may sound unremarkable, but it has been unknown for a long time whether this was possible. Part of the reason this is more difficult than it sounds is that cubes can be both positive and negative.

By contrast, it is very easy to decide whether it is possible to write 33 as a sum of two squares, $33 = a^2 + b^2$. Since squares can only be positive, any values of $a$ and $b$ greater than $5$ are not going to work, since they would make the sum too big. So there are only a few pairs of values to check: it’s enough to just check all pairs $(a,b)$ with $1 \leq a \leq b \leq 5$ (quick: how many such pairs are there?), which can even be done by hand in a few minutes. None of them work, so this exhaustive search of all the possibilities proves that it is not possible to find $a$ and $b$ such that $33 = a^2 + b^2$.

But a sum of three cubes, $33 = a^3 + b^3 + c^3$, is an entirely different matter! We can’t put any a priori bound on the size of the values $a$, $b$, and $c$, because the cube of a negative number is negative—if we choose at least one of them to be positive and at least one of them to be negative, they could in theory be very large but “cancel out” to yield 33. And that’s exactly what Dr. Booker found (using approximately 23 years’ worth of computer time spread over one month!): $33 = 8\,866\,128\,975\,287\,528^3 + (-8\,778\,405\,442\,862\,239)^3 + (-2\,736\,111\,468\,807\,040)^3.$

Whoa. 