Hypothesis K

Hardy and Littlewood [HaLi], in their investigations into Waring's problem via the circle method, made the following conjecture.
Hypothesis K
Let $k\geq 1$, and let $A$ be the set of non-negative $k$th powers. For any $n\geq 1$, \[1_A^{(k)}(n) \ll_\epsilon n^\epsilon.\]
This is trivially true when $k=1$, and follows from the classical theory of quadratic forms when $k=2$. It is an incredibly strong hypothesis - notice, for example, that it implies the $2k$-fold additive energy of the $k$th powers is essentially as small as it can be. Indeed, coupling this with the basic circle method is already enough to prove that every sufficiently large integer is the sum of $O(k)$ many $k$th powers. (The current bounds we have are of the form $O(k\log k)$.) Hardy and Littlewood [HaLi] write: "It is not too much to say that, if Hypothesis K were proved, Waring's Problem would be within measurable distance of a final solution."

Given a) it was known to be true for $k=1,2$ and b) it would have some very strong consequences and c) it seems true from a randomness heuristic point of view, if we replace the $k$th powers by a random set of the same density, it was reasonable for Hardy and Littlewood to highlight it as a special hypothesis, and to direct attention towards its investigation.

Unfortunately, it is false when $k=3$.

Mahler's disproof for $k=3$

In 1936 Mahler [Ma] disproved Hypothesis K when $k=3$, via an explicit algebraic relationship that shows that whenever $n$ is a 12th power it can be written as the sum of $\gg n^{1/12}$ positive cubes. The presentation of Mahler makes this algebraic discovery seem somewhat magical. The purpose of this note (which came out of a conversation with Sarah Peluse) is to explain how one might discover Mahler's relationship for oneself.

Our goal is discover some kind of parametric family of solutions to $x_1^3+x_2^3+x_3^3=n$, for some suitable family of $n$, where $x_i$ are restricted to positive integers. The key starting observation is that if we set $x_i=y_i-y_i'$ we can expand out the cubes and hope to cancel out most of the terms on the left-hand side, leaving just $n$. The simplest way of doing this is to study something of the form \[u^3 + (n^{1/3}-x)^3 + (y-u)^3 = n,\] where $u$ is some free parameter and $x,y$ are some functions of $n$ and $u$. The point is that the left-hand side naturally has an $n$, and the $u^3$ is cancelled automatically also. Therefore we just need to find some appropriate choice of $x,y$ to make the rest of the left-hand side cancel. That this is possible (while ensuring $u,x,y$ are all integers, and $x \leq n^{1/3}$ and $u\leq y$) is not at all guaranteed! Fortunately it will turn out to be so in quite a simple way. Expanding out the left-hand side gives \[n+3n^{1/3}x^2+y^3+3yu^2 - 3n^{2/3}x-x^3-3uy^2.\] The simplest way to make sure everything except $n$ vanishes here is to have them cancel pairwise. Let's try choosing $x,y$ such that $y^3=3n^{2/3}x$, so $x=y^3/3n^{2/3}$. This cancels one pair. Next we'll cancel $x^3$ with $3yu^2$, so $y=x^3/3u^2=y^9/3^4n^2u^2$, which forces $y=3^{1/2}n^{1/4}u^{1/4}$, and hence $x=3^{1/2}n^{1/12}u^{3/4}$.

Two pairs are cancelled, but we have no degrees of freedom left. Still, if we're very lucky, maybe the other terms will cancel anyway... we have a $3n^{1/3}x^2$ and a $3uy^2$. Fortunately, $n^{1/3}x^2=3n^{1/6+1/3}u^{3/2}=3n^{1/2}u^{3/2}=uy^2$, so we escape by the skin of our teeth, and have a representation of $n$ as the sum of three cubes.

We're basically done, we just need to massage things a little so that all three cubes are non-negative integers. For this we need $u,n^{1/3},3^{1/2}n^{1/4}u^{1/4},3^{1/2}n^{1/12}u^{3/4}$ to all be integers, and $u \leq 3^{1/2}n^{1/4}u^{1/4}$, and $3^{1/2}n^{1/12}u^{3/4} \leq n^{1/3}$.

It's easy to check that these conditions are all satisfied if we take $u=9t^4$ for some integer $t\geq 0$, and $n$ is a $12$th power satisfying $9t^3 \leq n^{1/4}$. In particular, there are $\gg n^{1/12}$ valid choices of $t$, all of which give a representation of $n$ as the sum of three cubes.

Further notes

Despite the (with hindsight) relative ease with which Hypothesis K can be disproven when $k=3$, it remains an open question for all $k\geq 4$. Perhaps there is some complicated algebraic relationship that remains undiscovered that would disprove it. Or, even better, perhaps it is true when $k\geq 4$! This would be lovely, but I personally doubt it.

Although Hypothesis K is false, hope remains - as I noted above, an immediate consequence of Hypothesis K is the fact that $E_{2k}(A_{\leq N}) \ll_\epsilon N^{1+\epsilon}$, if $A_{\leq N}$ is the set of $k$th powers in $[1,N]$. This is the estimate which actually matters for the circle method, and one might hope that this weaker claim (sometimes called Hypothesis K*) is true even when Hypothesis K fails.

And when $k=3$ this is true! This was proved (conditionally on the Generalised Riemann Hypothesis) independently by Hooley and Heath-Brown. As far as I know, this weaker hypothesis is still an open question when $k\geq 4$, however.