Hypothesis K

Let $k\geq 1$, and let $A$ be the set of non-negative $k$th powers. For any $n\geq 1$,
\[1_A^{(k)}(n) \ll_\epsilon n^\epsilon.\]
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 $k=3$.

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.

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.

**[Ma]**K. Mahler, Note on Hypothesis K of Hardy and Littlewood, J. London Math. Soc. 11 (1936), no. 2, 136-138.**[HaLi]**G. H. Hardy and J. E. Littlewood, Some problems of 'Partitio Numerorum' (VI): Further researches in Waring's Problem, Math. Z. (1925), 1-37.