Numbers as Sums of Two Rational Cubes

... one of the oldest questions in number theory: How many integers can be written as the sum of two cubed fractions, or, as mathematicians call them, rational numbers? The numbers 6 and 13, for example, can be written as: $$ \begin{align} 6&= \left ( \dfrac{17}{21} \right ) ^3 +  \left ( \dfrac{37}{21} \right  ) ^3\\ 13 &=  \left ( \dfrac{7}{3} \right ) ^3+ \left ( \dfrac{2}{3} \right )^3 \end{align}$$Mathematicians have suspected for decades that half of all integers can be written this way. Just as with odd and even numbers, this property appears to divide whole numbers into two equal camps: those that are the sum of two cubes, and those that aren’t.

I came across this statement in a Quanta article and it came as news to me that it was "one of the oldest questions in number theory". Well, at least now I know. My immediate thought was how to translate this problem into one involving just integers rather than fractions. It's not difficult. For an integer \(n\):$$ \begin{align} \text{if }  \left ( \frac{a}{c} \right ) ^3 + \left ( \frac{b}{c} \right  ) ^3&=n\\ \text{then }a^3+b^3&=c^3 \times n \end{align}$$Using this approach, it's easy enough to use SageMathCell to generate the result for say 13 (shown earlier). Here is the permalink. Figure 1 shows the numbers from 1 to 100 that can be written as a sum of two cubed fractions:

Figure 1: numbers in blue can be written as the sum of
two cubed rational numbers; the others cannot. 

Some of these numbers can be written as the sum of two integers cubed e.g. \(9=1^3+2^3\) or \(35=2^3+3^3\) but are of course equivalent to fractions when written as:$$ \begin{align} 9&= \left ( \dfrac{1}{1} \right ) ^3 +  \left ( \dfrac{2}{1} \right  ) ^3\\ 35 &=  \left ( \dfrac{2}{1} \right ) ^3+ \left ( \dfrac{3}{1} \right )^3 \end{align}$$The need for this clarification arises from the use of the imprecise term "fraction" whereas the term "rational number" should be used instead. Some of the pairs of rational numbers, whose cubes add to the blue numbers shown in Figure 1, are not easy to calculate. Take for instance:$$x^3+y^3=97$$The graph of this function is shown in Figure 2.

Figure 2: Geogebra

The values of \(x\) and \(y\) must be positive and lie between \(0\) and \(97^{1/3} \approx 4.5947\). Using SageMathCell, the calculation quickly times out and using my Jupyter notebook the calculation is still chugging away. It remains to be seen whether my algorithm will eventually display a solution. As the article says:

In the sum-of-two-cubes problem, the fractions involved can be enormous: The number 2,803, for example, is the sum of two cubed fractions whose denominators each have 40 digits.

The article goes on to say that:

Mathematicians have suspected for decades that half of all integers can be written this way (as a sum of the cubes of two rational numbers). Just as with odd and even numbers, this property appears to divide whole numbers into two equal camps: those that are the sum of two cubes, and those that aren’t. But nobody was able to prove this, or even give any bound on the proportion of whole numbers that fall into each camp. As far as mathematicians knew, the camp consisting of sums of rational cubes might be vanishingly small — or it might contain nearly every whole number. Mathematicians have calculated that, if something called the Birch and Swinnerton-Dyer conjecture is true (as is widely believed), about 59% of numbers up to 10 million are the sum of two rational cubes. But such data can, at best, offer hints about how the rest of the number line might behave.

What the paper discussed in this article found was that at least 2/21 (about 9.5%) and at most 5/6 (about 83%) of whole numbers can be written as the sum of two cubed fractions.

Getting back to these fractions, I discovered helpful resources at this site which assisted me in completing the following information about the sums of cubes up to 100. The list is still incomplete and I'll endeavour to complete it as time goes by. It's interesting that three different sums are given on the site for 19 and I'm wondering if the other numbers whose sums have been completed can also be represented in alternative ways.

\(1\)     pending
\(2\)     pending
\(6 = \left ( \dfrac{17}{21} \right )^3 + \left( \dfrac{37}{21} \right )^3 \) 
\(7 = \left ( \dfrac{5}{3} \right )^3 + \left( \dfrac{4}{3} \right )^3 \)
\(8\)    pending
\(9 = \left ( \dfrac{1}{1} \right )^3 + \left( \dfrac{2}{1} \right )^3 \)
\(12 = \left ( \dfrac{19}{39} \right )^3 + \left( \dfrac{89}{39} \right )^3 \)
\(13 = \left ( \dfrac{2}{3} \right )^3 + \left( \dfrac{7}{3} \right )^3 \)
\(15 = \left ( \dfrac{397}{294} \right )^3 + \left( \dfrac{683}{294} \right )^3 \)
\(16\)    pending
\(17\)    pending
\(19 = \left (\dfrac{1}{3} \right )^3 + \left (\dfrac{8}{3} \right )^3 = \left (\dfrac{5}{2} \right) ^3 + \left (\dfrac{3}{2} \right )^3 =\left ( \dfrac{92}{35} \right )^3 + \left (\dfrac{33}{35} \right )^3 \)
\(20 = \left ( \dfrac{1}{7} \right )^3 + \left( \dfrac{19}{7} \right )^3 \)
\(22\)    pending
\(26\)    pending
\(27\)    pending
\(28 = \left ( \dfrac{1}{1} \right )^3 + \left( \dfrac{3}{1} \right )^3 \)
\(30\)    pending
\(31\)    pending
\(33\)    pending
\(34\)    pending
\(35 = \left ( \dfrac{2}{1} \right )^3 + \left( \dfrac{3}{1} \right )^3 \)
\(37 = \left ( \dfrac{18}{7} \right )^3 + \left( \dfrac{19}{7} \right )^3 \)
\(42\)    pending
\(43 = \left ( \dfrac{1}{2} \right )^3 + \left( \dfrac{7}{2} \right )^3 \)
\(48 = \left ( \dfrac{34}{21} \right )^3 + \left( \dfrac{74}{21} \right )^3 \)
\(49\)    pending
\(50\)    pending
\(51\)    pending
\(53\)    pending
\(54\)    pending
\(56 = \left ( \dfrac{8}{3} \right )^3 + \left( \dfrac{10}{3} \right )^3 \)
\(58\)    pending
\(61\)    pending
\(62 = \left ( \dfrac{7}{3} \right )^3 + \left( \dfrac{11}{3} \right )^3 \)
\(63\)    pending
\(64\)    pending
\(65 = \left ( \dfrac{1}{1} \right )^3 + \left( \dfrac{4}{1} \right )^3 \)
\(67\)    pending
\(68\)    pending
\(69\)    pending
\(70 = \left ( \dfrac{17}{13} \right )^3 + \left( \dfrac{53}{13} \right )^3 \)
\(71\)    pending
\(72 = \left ( \dfrac{2}{1} \right )^3 + \left( \dfrac{4}{1} \right )^3 \)
\(75\)    pending
\(78\)    pending
\(79\)    pending
\(84\)    pending
\(85\)   pending
\(86 = \left ( \dfrac{5}{3} \right )^3 + \left( \dfrac{13}{3} \right )^3 \)
\(87\)    pending
\(89 = \left ( \dfrac{36}{13} \right )^3 + \left( \dfrac{53}{13} \right )^3 \)
\(90\)    pending
\(91 = \left ( \dfrac{3}{1} \right )^3 + \left( \dfrac{4}{1} \right )^3 \)
\(92\)    pending
\(94\)    pending
\(96 = \left ( \dfrac{38}{39} \right )^3 + \left( \dfrac{178}{39} \right )^3 \)
\(97\)    pending
\(98 = \left ( \dfrac{355}{152} \right )^3 + \left( \dfrac{669}{152} \right )^3 \)