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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: $$4104 = 2^3+16^3 = 9^3+15^3,$$ $$13832 = 20^3+18^3=24^3+2^3,$$ $$20683 = 10^3 +27^3 = 19^3 +24^3.$$

Community
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Mankind
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    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. – MJD May 09 '15 at 22:37
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    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, …. – MJD May 09 '15 at 22:40
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    FYI : OEIS A001235 – mathlove May 09 '15 at 22:43
  • @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. – Mankind May 09 '15 at 22:50
  • @MJD thanks, I have included some of your findings in my post. – Mankind May 09 '15 at 22:55
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    Here's the first thousand examples. – MJD May 09 '15 at 23:47
  • @HowDoIMath: See Leonhard Euler, Disquitiones Artithmeticae, Vol. I, Ch. $272$, Pag. $556-576$. – Lucian Feb 23 '16 at 08:17
  • Just to add, there are also solutions where $a^3 - b^3$ can be the sum of two cubes in two different ways, for $a, b \in \mathbb{Z}^+$. $$46^3 - 37^3 = 46683 = 27^3 + 30^3 = 3^3 + 36^3$$ $$87^3 - 79^3 = 165464 = 38^3 + 48^3 = 20^3 + 54^3$$ $$202^3 - 171^3 = 3242197 = 85^3 + 138^3 = 76^3 + 141^3$$ I thought it would be worth mentioning :) – Mr Pie Nov 14 '17 at 12:01
  • And $229^3 - 192^3 = 4931101 = 102^3 + 157^3 = 76^3 + 165^3$ and another $256^3 - 255^3 = 195841 = 22^3 + 57^3 = 9^3 + 58^3$ – Mr Pie Nov 14 '17 at 22:08
  • And $419^3 - 362^3 = 26122131 = 235^3 + 236^3 = 107^3 + 292^3$ and also $391^3 - 382^3 = 4033503 = 87^3 + 150^3 = 24^3 + 159^3$ and just one more, $453^3 - 397^3 = 30388904 = 234^3 + 260^3 = 26^3 + 312^3$ – Mr Pie Nov 14 '17 at 22:16
  • Oh, and $517^3 - 436^3 = 55306557 = 285^3 + 318^3 = 6^3 + 381^3$. There are infinitely many – Mr Pie Nov 14 '17 at 22:22

3 Answers3

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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 $m$.)

Tito Piezas III
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  • 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. – Mankind May 10 '15 at 17:05
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    @HowDoIMath: I found Broughan's theorem that enabled me to test $F(4),\dots,F(60)$. Now it is even more promising. – Tito Piezas III May 11 '15 at 04:47
  • 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. :) – Mankind May 12 '15 at 18:47
  • @TitoPiezasIII: See Leonhard Euler, Disquitiones Artithmeticae, Vol. I, Ch. $272$, Pag. $556-576$. – Lucian Feb 23 '16 at 08:20
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    This holds for $N\le 200$. I'm not clear on how you came up with this formula, though, even given the paper quoted. – rogerl May 26 '16 at 18:08
  • @rogerl: You mean $F(n)$? I've now forgotten where I found this. I re-checked Broughan's paper and it's not from there. Hm... – Tito Piezas III Jan 16 '18 at 17:13
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As for the question regarding whether or not there are infinitely many numbers that can be expressed as the sum of two cubes in two different ways or not, there's a very quick and simple way to prove this. Since you have $$ 1729 = 10^3 + 9^3 = 12^3 + 1^3 $$ Multiply both sides by $ n^3 $ where $ n $ is a positive integer to get $$ 1729 n^3 = (10n)^3 + (9n)^3 = (12n)^3 + n^3 $$ and there you have it, plug in any value for $ n $ and and you have your proof for the infinitude of such numbers. Obviously this doesn't include all the solutions but it proves that there certainly are an infinite number of them

Jibran
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    It is obvious that such numbers have at least two representations as a sum of two cubes, but is it obvious that they have exactly two? – Daniel Fischer Oct 06 '19 at 14:38
  • No, well not to me it isn't. – Jibran Oct 06 '19 at 18:30
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    If $(10n)^3 + (9n)^3 = (an + b)^3 + (cn + d)^3$ then $b^3 + d^3$ must be 0 modulo $n$, and if $d$ is invertible then $bd^{-1}$ must be a cube root of $-1$ modulo $n$. So you can choose $n$ from the primes of the form $3k + 2$, of which there are infinitely many, then there is no such cube root of $-1$, so $b$ and $d$ must be multiples of $n$, and the result follows. – kaya3 Feb 10 '20 at 04:54
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I found a number recently which can be expressed as a sum of two cubes in exactly two different ways.

  65673928=164³+394³=103³+401³
SHOUNAK GUPTA
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