Sums of two cubes

How many numbers can be represented as sums of two cubes? How many of them in exactly one way? How many of them in at least ways etc?

Counting the the number of points $x^3+y^3 \le X$ we count the numbers with multiplicity and it turns out to be $\sim CX^{2/3}$ . (Each of the variables has around $X^{1/3}$ choices which shows you the order of magnitude, and to count more precisely, you can count lattice point inside the curve $x^3+y^3 =X$ .
What if we want to count without multiplicity? The above is the sum $\sum_{n \le X} r(n)$ where $r(n)$ is the number of representation of $n$ as sum of two cubes. We want to be able to get estimates on quantities like $\sum_{n \le X} r(n)^k$ , for large $k$ , to bound the contribution from high multiplicity. Once we have that information, we can show that the number without multiplicity is $\sim \frac{C}{2}X^{2/3}$ . Each number if it’s counted has at least two representations by reversing orders $(x,y)$ and $(y,x)$ . And this says that the contribution of numbers with genuine multiple representations (not just switching order) is neglible. In fact the following bound is true.

The size of numbers with multiple representation as sums of cubes is $O(x^{5/9+\varepsilon})$ This is a result due to Christopher Hooley and a much simpler proof is given by Trevor Wooley. Both of them try to count $x^3+y^3 =r^3+s^3 \le X.$

After some change of variables, Hooley’s method involves using bounds on representation number binary quadratic forms in some cases; In another harder case, he uses a sieve to detect the condition that one of the variable is a square, thus reduces counts to arithmetic progressions. To deal with the error terms from arithmetic progressions, he using additive characters to detect progressions and then the error terms can be bounded in terms of exponential sums of the form $\displaystyle \sum_{X^3+Y^3 =1+Z^2 \mod p} e_p(aX+bY)$ .

Wooley does it by count the solutions by fixing $h =x+y -r-s.$ Then after a change of variables, the problems reduces to counting solution for $X^2+DY^2 =n$ for a fixed $n$ .