Deep thought required

The integer 10 can be written as the sum of three cubes: $10=2^3+1^3+1^3$. So can 6, although that isn’t obvious until you think of using negative integers: $6 = 2^3+(-1)^3+(-1)^3$. For that matter, there are easy solutions to the problem of finding such sums of three cubes for every integer up to 10 except for 4 and 5:

• $0 = 0^3+0^3+0^3$
• $1 = 1^3+0^3+0^3$
• $2 = 1^3+1^3+0^3$
• $3 = 1^3+1^3+1^3$
• $4 =$???
• $5 =$???
• $6 = 2^3+(-1)^3+(-1)^3$
• $7 = 2^3+(-1)^3+0^3$
• $8 = 2^3+0^3+0^3$
• $9 = 2^3+1^3+0^3$
• $10 = 2^3+1^3+1^3$

That prompts a few questions: Are there solutions for 4 and 5? Are there solutions for all integers? Or for an infinite number of integers?

The answers to the first two questions are, no and no. Start by observing that the cube of any multiple of 3, $(3n)^3$, is divisible by 9 (in fact 27). For numbers equal to 1 mod 3, $(3n+1)^3 = 27n^3+27n^2+9n+1$ so is equal to 1 mod 9, and for numbers equal to -1 mod 3, $(3n-1)^3 = 27n^3-27n^2+9n-1$ so is equal to -1 mod 9. So all cubes are equal to 0 or ±1 mod 9. From this you can figure out all sums of three cubes are equal to 0, 1, 2, or 3 mod 9. But 4 and 5 equal -4 and +4 mod 9. So they cannot be written as sums of three cubes; nor can 13, 14, 22, 23, 31, 32, ….

As for the third question, it’s conjectured all other integers can be expressed as sums of three cubes. But it’s not proved.

The integers up through 11 are easy. In fact you might realize 10 can be done another way: $10 = 4^3+(-3)^3+(-3)^3$. For that matter, $8 = 2^3+1^3+(-1)^3 = 2^3+2^3+(-2)^3$ and so on for an infinite number of solutions.

But 12 might take you a little longer: $12 = 10^3+7^3+(-11)^3$. Worse, $21 = 16^3+(-11)^3+(-14)^3$. Things are pretty easy up through 29, and then 30 is hard. Very hard.

Good luck coming up with the simplest known answer:

$30 = 2220422932^3+(-283059965)^3+(-2218888517)^3$.

Yikes!

So it goes. Lots of easy ones

$53 = 3^3+3^3+(-1)^3$

interspersed with slightly tougher ones

$52 = 60702901317^3 + 23961292454^3 + (-61922712865)^3$

But as of a few days ago, there were solutions known for all but two numbers less than 100 (other than the ±4 mod 9 impossibilities).

That was then. Andrew R. Booker has just published this result for the smallest hitherto-unsolved case:

$8866128975287528^3 + (-8778405442862239)^3 + (-2736111468807040)^3$

$= 696950821015779435648178972565490929714876221952$
$-676467453392982277424361019810585360331722557919$
$-20483367622797158223817952754905569383153664000$

$= 696950821015779435648178972565490929714876221952$
$-696950821015779435648178972565490929714876221919$

$= 33$

I hardly need add that this was not found with a Python script during Booker’s lunch hour. The paper goes into the number theoretic gymnastics involved in his computer search, which, he reports, “used approximately 15 core-years over three weeks of real time.”

So that’s 33 solved. We’re down to one remaining unsolved number under 100. And I’m ready to give it to you. Now.

Though I don’t think that you’re going to like it.

All right.

You’re really not going to like it.

All right. The smallest $n \ne \pm4\mod 9$ for which no expression as the sum of three cubes is known is…

is…

Forty-two.