1
$\begingroup$

A Taxi Cab Number is a number $\text{Ta}(n)$ that can be written as the sum of two cubes in $n$ different ways. More formally, they are known as Hardy-Ramanujan Numbers or, as Ramanujan had called them, Magic Numbers.


$$\begin{align} 577^3 &= 356^3 + 385^3 + 448^3 \\ &= 1^3 + 426^3 + 486^3 \\ &= 172^3 + 318^3 + 537^3 \\ &= 41^3 + 244^3 + 562^3 \\ &= 153^3 + 174^3 + 568^3 \\ &= 90^3 + 201^3 + 568^3 \end{align}$$ Is this the smallest cube that is expressible as the sum of $3$ positive cubes in $6$ different ways? In symbols, does $$577^3 = \text{taxicab}(3, 3, 6)\,?\tag*{$\bigg(\begin{align} \verb|S|&\verb|uch that the LHS| \\ &\verb|must be a cube.|\end{align}\bigg)$}$$ I have been trying to find numbers similar to Taxi Cab Numbers. The best example of a taxicab number is $1729$ because it is the smallest number that can be expressed as the sum of two cubes in two different ways, i.e. $$1729 = 1^3 + 12^3 = 9^3 + 10^3.$$ I developed a more general case of finding numbers $a_n^{ \ \ 2n - 1}$ that can be written as the sum of $(n+1)$ cubes in $2(2n-1)$ different ways, trying to find a potential pattern in the values of $a_n$ for which $a_n \in\mathbb{Z^+}$.

Given that $n = 1$, we obtain that $a_1 = 1729$. Given $n = 2$, we apparently obtain $a_2 = 577$. If this is true, the only pattern I can find is that $$a_n^{\ \ 2n - 1} - 1 = \left\{\sum_{k=1}^n b_k^{\ \ 3} : b_k\in\mathbb{Z^+}\right\}.$$ Also, could somebody find the value of $a_3$ and $a_4$ and if they want, $a_5$ (though I think the smallest number that is the sum of $9$ cubes in $18$ different ways might be pretty large).

Thank you in advance.


P.S. I was not able to find any other appropriate tags apart from > (number-theory) <

$\endgroup$
5
  • $\begingroup$ Is this the smallest cube that is expressible as the sum of $3$ cubes in $6$ different ways? In symbols, does $$577^3 = \text{taxicab}(3, 3, 6)\,?$$ This is not exactly the same question: taxicab(3,3,6) may not be a cube itself. $\endgroup$ Commented Feb 25, 2018 at 11:09
  • $\begingroup$ @ArnaudMortier true that. Perhaps I will make it clearer. Thanks for pointing that out :) $\endgroup$ Commented Feb 25, 2018 at 11:21
  • 2
    $\begingroup$ $577^3$ is certainly not the smallest cube that can be expressed as a sum of three cubes in six different ways, as $1^3$ can be expressed as a sum of three cubes in infinitely many different ways – see, e.g., math.stackexchange.com/questions/32559/… But perhaps you meant a sum of three positive cubes. If so, please edit your question to reflect that intention. $\endgroup$ Commented Feb 25, 2018 at 11:57
  • $\begingroup$ @GerryMyerson yes I did, because I knew of the one with $1^3$ so thank you for reminding me :) $\endgroup$ Commented Feb 25, 2018 at 12:42
  • $\begingroup$ @GerryMyerson I commented here $\longrightarrow$ mathoverflow.net/questions/138886/… $\endgroup$ Commented Feb 25, 2018 at 12:51

1 Answer 1

4
$\begingroup$

No.

$$\begin{align}216^3 &=24^3+ 144^3+ 192^3\\ &=26^3 +102^3+208^3\\ &=30^3 +164^3+ 178^3\\ &=48^3 +76^3+ 212^3\\ &= 102^3 +117^3 +195^3\\ &=108^3 +144^3+ 180^3\end{align}$$

Find more here.

$\endgroup$
5
  • $\begingroup$ Thank you very much! But just out of curiosity, if $a_2^{\ \ 3}$ had to strictly be a cube number, and not like $216^3$ because that is equal to $6^9$ as well, then would $a_2 = 577$? I cannot vote now because I have reached my daily limit, so I have to wait a couple of hours. $\endgroup$ Commented Feb 25, 2018 at 12:45
  • 1
    $\begingroup$ I don't know, but I think a little script in Mathematica testing only strict cubes could answer that question. I can only use the online version so can't run For loops. $\endgroup$ Commented Feb 25, 2018 at 12:50
  • $\begingroup$ Thank you for that :) ..... But how do I get Mathematica on my computer? Is there an online version? $\endgroup$ Commented Feb 25, 2018 at 12:53
  • 1
    $\begingroup$ Wolfram Alpha is the online version. $\endgroup$ Commented Feb 25, 2018 at 17:30
  • $\begingroup$ Thank you. I will look into it :) $\endgroup$ Commented Feb 25, 2018 at 17:34

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.