What is the taxicab number for rational fourth powers?

Question

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.

Answer 1

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.

Answer 2

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.