16
$\begingroup$

The taxicab number is the smallest integer that can be expressed as a sum of two positive integer cubes in two different ways, and it is equal to $1729=12^3+1^3=10^3+9^3$. There are generalizations to more than two ways, to allowing negative cubes (then the answer is $91=4^3+3^3=6^3-5^3$), and to different powers. In the latter case, the answer is $50=7^2+1^2=5^2+5^2$ for squares, $635318657 = 134^4 + 133^4 = 158^4 + 59^4$ for fourth powers, while for fifth powers it is not known whether any such number exists, that is, whether equation $x^5+y^5=z^5+t^5$ has any non-obvious integer solutions.

It is natural to ask the same question for rational powers, that is, what is the smallest positive integer that can be expressed as a sum of two rational $d$-th powers in two different ways. For squares, the answer is $1=(3/5)^2+(4/5)^2=(5/13)^2+(12/13)^2$, see a comment of Wojowu. For cubes, the answer is $$ 6 = \left(\frac{37}{21}\right)^3 + \left(\frac{17}{21}\right)^3 = \left(\frac{2237723}{960540}\right)^3 + \left(-\frac{1805723}{960540}\right)^3. $$ In fact, $6$ has infinitely many such representations because $x^3+y^3=6z^3$ defines an elliptic curve of rank $1$. For fifth powers, no such positive integers are known, because if they were, we could obtain non-trivial integer solution to $x^5+y^5=z^5+t^5$ after multiplying by denominators.

This leaves the case of $d=4$. That is: what is the smallest positive integer that can be expressed as a sum of two rational fourth powers in two different ways?

The sequence of positive integers $n$ that are the sums of two rational fourth powers starts with $1,2,16,17,32,81,82,\dots$. From the solutions to well-known equations $x^4+y^4=z^4$ and $x^4+y^4=2z^4$ we conclude that such representations for $1,2,16,32,81$ are unique. Jeremy Rouse informed me that in 2001, Flynn and Wetherell proved that the representations of $17$ is unique as well. Hence, the first case for which I currently do not know the answer is $n=82$.

The challenge is that the existence of one representation implies that some standard methods (such as local obstructions, reduction to rank $0$ elliptic curves, or the Mordell-Weil Sieve in its simplest form) does not help for proving the non-existence of a second representation.

$\endgroup$
10
  • $\begingroup$ For squares, the answer is $1=(3/5)^2+(4/5)^2=(5/13)^2+(12/13)^2$. Any nonzero rational which is a sum of two rational squares can be written as such a sum in infinitely many ways. $\endgroup$
    – Wojowu
    Commented Sep 2, 2023 at 11:38
  • $\begingroup$ Thank you, I have updated this with reference to your comment. $\endgroup$ Commented Sep 2, 2023 at 11:43
  • $\begingroup$ $17$ has another representation in terms of not equal powers $17=2^3+3^2$. $\endgroup$
    – user25406
    Commented Sep 2, 2023 at 20:08
  • 5
    $\begingroup$ I should have consulted Prof. Google before posting. proofwiki.org/wiki/… says $5906=(149/17)^4+(25/17)^4$ is the smallest integer sum of two rational 4th powers that is not a sum of two integer 4th powers. $\endgroup$ Commented Sep 2, 2023 at 23:22
  • 1
    $\begingroup$ @GerryMyerson with denominators of $41$ the smallest case seems to be $(1002/41)^4+(751/41)^4=469297$. The minimum for a given denominator is growing rapidly with said denominator and this augurs against fractional solutions smaller than the known integer minimum. $\endgroup$ Commented Feb 28 at 1:47

2 Answers 2

16
$\begingroup$

It has been known since Euler that the quartic surface defined by

$$\displaystyle x_1^4 + x_2^4 = x_3^4 + x_4^4$$

contains a rational curve defined by

\begin{gather*} \displaystyle x_1(t) = t^7 + t^5 - 2t^3 + 3t^2 + t, \\ \displaystyle x_2(t) = t^6 - 3t^5 - 2t^4 + t^2 + 1, \\ \displaystyle x_3(t) = t^7 + t^5 - 2t^3 - 3t^2 + t, \\ \displaystyle x_4(t) = t^6 + 3t^5 - 2t^4 + t^2 + 1. \end{gather*}

This always corresponds to an integer point on the surface, and in particular, when $t \in \mathbb{Z}$ and $|t|\ge2$ this corresponds to a number with taxicab number at least two.

For $n \geq 5$, it has been a long-standing conjecture that for every integer $m$ such that $x^n + y^n = m$ has a solution in positive integers $x,y$, the only solutions are $(x,y)$ and $(y,x)$. This is a special case of a conjecture of C.L. Stewart, which is itself a special case of the uniform boundedness conjecture for algebraic curves.

Reference:

C.L. Stewart, On the number of solutions of polynomial congruences and Thue equations, J. Amer. Math. Soc. 4 (1991), 793–835.

$\endgroup$
2
  • 3
    $\begingroup$ Thank you for relevant discussion and a reference. However, unless I am missing something, this does not help to answer what is the taxicab number for rational fourth powers, and in particular is it equal to 17 or not. $\endgroup$ Commented Sep 2, 2023 at 18:11
  • 1
    $\begingroup$ Technically, we do not get two nontrivially different representations for all $t\in\mathbb{Z}$. The fourth powers of $x_3,x_4$ could match those if $x_1,x_2$, and so they end up doing for $t\in\{-1,0,1\}$. The claim should include the constraint $|t|\ge2$. $\endgroup$ Commented Sep 2, 2023 at 18:59
2
$\begingroup$

One approach is to seek numbers $N$ having the form

$$N=a^4+b^4=\dfrac{c^4+d^4}{e^4}$$

where all variables are $\in\mathbb{Z}_{>0}$ and $e>1$. Since $e$ must have only prime factors of the form $8n+1$, the smallest available value for that parameter is $17$. Then $c^4\equiv -d^4\bmod 17^4$ from which we may take

$$d\equiv\pm20051c.$$

(We also have roots given by $c\equiv\pm20051d$, but these do not give additional fourth-power sums because $c$ is interchangeable with $d$.) Searching through all possibile ordered pairs $(c,d)$ satisfying this constraint and giving $N<635318657$ yields no solutions in which $a$ and $b$ are also whole numbers (except a few cases where $c$ and $d$ are divisible by $e=17$ and we recover integer powers). Thus if any smaller "rational-base taxicab" numbers exist with exponent $4$, we must use bases having denominators larger than $17$.

Update, February 2024:

A similar method has been applied to the proposed equations

\begin{gather*} N={a^4+b^4}=\dfrac{c^4+d^4}{41^4} \\ N=\dfrac{a^4+b^4}{17^4}=\dfrac{c^4+d^4}{41^4}. \end{gather*}

Again no solutions with $N<635318657$ are found, so denominators larger than $41$ would be needed.

There were much fewer candidate sums with a denominator of $41$ in the base than with a denominator of $17$, so a conjecture is made that there are no fractional solutions at all with $N<635318657$. Then $635318657$ would become the minimal fourth-power taxicab number allowing all rational bases.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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