# Two, three, four

Last year was a big year for the sum of three cubes problem, which asks whether all numbers not of the form $9k+4$ or $9k+5$ can be written as the sum of the cubes of three (positive, negative, or zero) integers, $a^3+b^3+c^3$. It’s a hard problem. The smallest solution for 42 is

$42 = (-80\ 538\ 738\ 812\ 075\ 974)^3 + 80\ 435\ 758\ 145\ 817\ 515^3 + 12\ 602\ 123\ 297\ 335\ 631^3$

A superficially similar problem is to find ways to write an integer $N$ as the sum of a square, a cube, and a fourth power: $N = a^2+b^3+c^4$. As with the three cubes problem, we allow $b$ to be negative, though obviously there’s nothing to be gained by considering negative values for $a$ or $c$.

It’s not at all obvious all numbers can be so written, or that they can’t, or which numbers can’t if there are any, or whether it’s harder or easier to find such representations than it is to find sums of three cubes. Of course any number that is itself a square, cube, or fourth power has a solution using zero for the other two powers, and any number that is one or two more than a square, a cube, or a fourth power is trivial as well, as is any number that is one less than a square or a fourth power. And compared to the three cubes problem, the search space here is in some sense smaller since we require $0 \le a^2 \le (N-b^3)$ and $0 \le c^4 \le (N-b^3)$.

In fact all positive numbers up to 100 except for one have solutions with relatively small numbers: the worst of them, in the sense of having a large minimum value for $W = a^2+|b|^3+c^4$, is $75 = 171^2+(-31)^3+5^4$. The exception is this one:

$67 = (213\ 662)^2+(-6\ 337)^3+676^4$

It’s reminiscent of the three cubes problem in that you have this much harder number embedded among much easier ones. But it’s nowhere near as hard as 42 as a sum of three cubes.

For the numbers -10 000 through 10 000 a brute force Python script quickly turns up solutions for all. The one with largest $W$ (3 times the next largest) is $-8\ 357 = (3\ 303\ 433)^2 + (-22\ 183)^3 + 239^4$.

Is there a proof all numbers have solutions? Maybe. There is an easy (to follow) proof that every integer is a sum of two squares and a cube of an integer, and a proof (paywalled) that almost all positive integers are sums of a square, a positive cube, and a fourth power. Not having shelled out to read it — the paper’s ten pages long, so I infer it’s not an easy proof — I don’t know if it identifies all the exceptions; if it does, then finding solutions with negative $b$ for those would prove our puzzle for positive $N$. In 71 years you’d think it would have been done by now, and perhaps extended to negative $N$, and maybe it has. But that’s all I’ve found so far. Finally, for expressing $N$ as $a^2+b^3+c^3$, this paper reports no proof but there are solutions for all $-4\ 000\ 000 \le N \le 2\ 000\ 000$.