It seems known that there are infinitely many numbers that can be expressed as a sum of two positive cubes in at least two different ways (per the answer to this post: Number Theory Taxicab Number).

We know that

$$1729 = 10^3+9^3 = 12^3 + 1^3,$$

and I am wondering if there are infinitely many numbers like this that can be expressed as the sum of two positive cubes in exactly two ways?

In fact, are there even any other such numbers?

EDIT: As provided by MJD in the comments section, here are other examples: $$4104 = 2^3+16^3 = 9^3+15^3,$$ $$13832 = 20^3+18^3=24^3+2^3,$$ $$20683 = 10^3 +27^3 = 19^3 +24^3.$$

  • 1
    $\begingroup$ A computer search should quickly find more examples. Because the roots are required to be positive, if you find that 192837465 (or whatever) is the sum of cubes in two ways, you only need to examines the sums of cubes of numbers up to $\sqrt[3]{192837465}$ to verify that no other pairs add up to 192837465. This is easy. $\endgroup$ – MJD May 9 '15 at 22:37
  • 2
    $\begingroup$ The computer instantly finds further examples $4104 = 2^3+16^3 = 9^3+15^3, 13832 = 8\cdot 1729, 20683 = 10^3+27^3 = 19^3+24^3$, 32832, 39312, 40033, …. $\endgroup$ – MJD May 9 '15 at 22:40
  • 1
    $\begingroup$ FYI : OEIS A001235 $\endgroup$ – mathlove May 9 '15 at 22:43
  • $\begingroup$ @mathlove thanks, though that is a list of numbers that has at least two representations as a sum of two cubes. I see that some of MJD's examples are on that list as well. $\endgroup$ – Mankind May 9 '15 at 22:50
  • 4
    $\begingroup$ Here's the first thousand examples. $\endgroup$ – MJD May 9 '15 at 23:47

In the paper Characterizing the Sum of Two Cubes, Kevin Broughan gives the relevant theorem,

Theorem: Let $N$ be a positive integer. Then the equation $N = x^3 + y^3$ has a solution in positive integers $x,y$ if and only if the following conditions are satisfied:

  1. There exists a divisor $m|N$ with $N^{1/3}\leq m \leq (4N)^{1/3}.$

  2. And $\sqrt{m^2-4\frac{m^2-N/m}{3}}$ is an integer.

The sequence of integers $F(n)$,

$$\begin{aligned} F(n) &= a^3+b^3 = (2 n + 6 n^2 + 6 n^3 + n^4)^3 + (n + 3 n^2 + 3 n^3 + 2 n^4)^3\\ &= c^3+d^3 = (1 + 4 n + 6 n^2 + 5 n^3 + 2 n^4)^3 + (-1 - 4 n - 6 n^2 - 2 n^3 + n^4)^3 \end{aligned}$$

for integer $n>3$ apparently is expressible as a sum of two positive integer cubes in exactly and only two ways.

$$\begin{aligned} F(4) &= 744^3+756^3 = 945^3+15^3\\ F(5) &= 1535^3+1705^3 = 2046^3+204^3\\ &\;\vdots\\ F(60) &=14277720^3+26578860^3 = 27021841^3+12506159^3 \end{aligned}$$

Using Broughan's theorem, I have tested $F(n)$ from $n=4-60$ and, per $n$, it has only two solutions $m$, implying in that range it is a sum of two cubes in only two ways. Can somebody with a faster computer and better code test it for a higher range and see when (if ever) the proposed statement breaks down? Incidentally, we have the nice relations,

$$a+b = 3n(n+1)^3$$

$$c+d = 3n^3(n+1)$$

Note: $F(60)$ is already much beyond the range of taxicab $T_3$ which is the smallest number that is the sum of two positive integer cubes in three ways.

$$T_3 \approx 444.01^3 = 167^3+436^3 = 228^3+423^3 = 255^3+414^3$$

(Using the theorem, this yields 3 values for $m$.)

  • $\begingroup$ I don't know how you came up with this, but it looks great. I know you have to be careful with these things, seeing as you for small numbers don't have a lot of cubes to choose from, but being true for all of $F(4),\ldots,F(26)$ sounds promising. $\endgroup$ – Mankind May 10 '15 at 17:05
  • 1
    $\begingroup$ @HowDoIMath: I found Broughan's theorem that enabled me to test $F(4),\dots,F(60)$. Now it is even more promising. $\endgroup$ – Tito Piezas III May 11 '15 at 4:47
  • $\begingroup$ This is a nice find! As far as I can tell from having skimmed the proof in Broughan's paper, then the $m$ in the two decompositions of $F(n)$ as a sum of two cubes correspond to $m_1=3n^4+9n^3+9n^2+3n$ and to $m_2=3n^3+3n^4$ (here $m=u+v$, where $F(n)=u^3+v^3$). By the theorem, $\sqrt{m^2-4\frac{m^2-F(n)/m}{3}}$ is integer for $m=m_i$, $i=1,2$, The real question is then if $\sqrt{m^2-4\frac{m^2-F(n)/m}{3}}$ is integer for any other values of $m$, such that $m|F(n)$. This looks like a non-trivial problem though. :) $\endgroup$ – Mankind May 12 '15 at 18:47
  • $\begingroup$ @TitoPiezasIII: See Leonhard Euler, Disquitiones Artithmeticae, Vol. I, Ch. $272$, Pag. $556-576$. $\endgroup$ – Lucian Feb 23 '16 at 8:20
  • $\begingroup$ This holds for $N\le 200$. I'm not clear on how you came up with this formula, though, even given the paper quoted. $\endgroup$ – rogerl May 26 '16 at 18:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.