The diophantine $a^{11} + b^{11} = c^{11} + d^{11} = s$ for Eisenstein integers.

Question

Let $a,b,c,d,s$ be distinct nonzero Eisenstein integers.

Claim : $$a^{11} + b^{11} = c^{11} + d^{11} = s$$

has only a finite number of solutions for any given $s$.

In fact at most $22$ for any $s$.

Is this true ? Is it known or studied ?

Any references ?

@ArielFishbein oh $2*11$ , where $11$ approximates 11th roots of unity but scaled ( multiplied by large real ) and approximated by Eisenstein integers. And $2$ the ways to arrange the sum. That is not formal ofcourse. Good question — mick, Dec 05 '23 at 23:43
So we fix an integer $s$ in the cyclotomic field $\Bbb Q(\omega)$, $\omega$ being a primitive third root of the unit. And ask for the solutions $(a,b)$ of $a^{11}+b^{11}=s$. (We may add $a\ne b$.) The question is only about the number of the solutions $(a,b)$ (finitely many), or we need also $(c,d)$ and that $22$ is part of the question? — dan_fulea, Dec 06 '23 at 00:03
@dan_fulea if we do not need the $(c,d)$ the question is different , we just add $a^{11}$ and $b^{11}$ and get a solution. But that is not the diophantine with $a,b,c,d$. So we DO NEED the c and d — mick, Dec 06 '23 at 00:20
I thought $s$ is given, fixed. And for such an $s$ we are searching for pairs $(a,b)$, identified with $(b,a)$, realizing $s$ as $a^{11}+b^{11}$. The question is thus to find numbers $s$ which can be realized in at least two different ways, and it is a question about the finiteness of the set of these $s$-values? — dan_fulea, Dec 06 '23 at 00:30
Are we counting $(a,b)$ pairs such that $a^{11}+b^{11}=s$? Or quads $(a,b,c,d)$ such that $a^{11}+b^{11}=c^{11}+d^{11}=s$? Can you give an example of $s$ such that there are 22 solutions? — Jyrki Lahtonen, Dec 06 '23 at 06:36
With $s=0$ we trivially have infinitely many solutions from $b=-a$, but that's not very interesting :-) — Jyrki Lahtonen, Dec 06 '23 at 07:05
@dan_fulea I myself am uncertain what the OP wants. But there is only one known solution which I gave in my answer, though I had to go beyond Eisenstein integers. — Tito Piezas III, Dec 06 '23 at 12:06
@JyrkiLahtonen There is only one known solution (not in Eisenstein integers), and I am not sure how he can extract 22 solutions from it. — Tito Piezas III, Dec 06 '23 at 12:10
@JyrkiLahtonen we are counting quads. But the difference between those quads and pairs is almost identical. If we can write it as a pair in 3 ways then by simple combinatorics we can write it as quads in 3 ways. — mick, Dec 06 '23 at 12:23
If no case of 22 solutions, then what kind of example data have you available, Mick? — Jyrki Lahtonen, Dec 06 '23 at 13:41
FWIW All the repetitions among the numbers $a^{11}+b^{11}$ with $a,b$ non-zero Eisenstein's integers drawn from the set $J={m+n\omega\mid |m|,|n|\le6}$ have either $b=-a$ or just intechange $a$ and $b$. In other words, no non-trivial repetitions in $J$. I included the solutions of the form $a=0, b=-z, c=\omega z, d=z\omega^2$ among the trivial ones (see my comment under Tito's answer). I guess I could easily extend the range, but I'm sure we have other users who are more capable programmers :-) — Jyrki Lahtonen, Dec 06 '23 at 16:17
@JyrkiLahtonen I have no nontrivial solutions. I am not even sure nontrivial solutions exist. I guess it is only like up to unit multiples and such. — mick, Dec 06 '23 at 19:21

Tito Piezas III · Answer 1 · 2023-12-06T12:13:02.480

4

The equation,

$$a^{11}+b^{11} = c^{11}+d^{11}\tag1$$

is difficult to solve using quadratic integers, especially if we limit ourselves to Eisenstein integers. So we generalize things slightly and consider the ring of quadratic integers of $\mathbb{Q}(\sqrt{D})$ that is both a "principal ideal domain" and "Euclidean domain". Then for negative discriminants $D$ we only have five:

$$D = (-1,\,-2,\,-3,\,-7,\,-11)$$

and, voila, we have the single known identity for $11$th powers,

$$\left(\frac{1+\sqrt{-7}}2\right)^{11}+\left(\frac{1-\sqrt{-7}}2\right)^{11}=\left(\frac{1+\sqrt{-11}}2\right)^{11}+\left(\frac{1-\sqrt{-11}}2\right)^{11}= 67$$

given in this 2017 post. I don't know of any other solution for $(1)$ using quadratic integers, nor if,

$$a^{13}+b^{13} = c^{13}+d^{13}\tag2$$

is also solvable, but is probably not.

edited Dec 06 '23 at 12:13

answered Dec 06 '23 at 11:32

Tito Piezas III

54,565

+1 though those are 2 distinct rings and not Eisenstein. – mick Dec 06 '23 at 12:18
@mick Yes. I've looked at this problem before. As it is of high degree, it may need two Euclidean domains, and not necessarily Eisenstein. Though I just realized that $$\big(2+\sqrt{-3}\big)^{11}+\big(2-\sqrt{-3}\big)^{11} = 2^2\cdot67$$ so the prime $p=67$ seems to be well-represented. – Tito Piezas III Dec 06 '23 at 12:26
I am probably misunderstanding your statement, but I think there is no shortage of solutions in $\mathcal{O}=\Bbb{Z}[\omega],\omega=e^{2\pi i/3}$. If $z\in\mathcal{O}$ is any Eisensteinian integer we have trivially $$z^{11}+(\omega z)^{11}+(\omega^2z)^{11}=0.$$ Yielding solutions of the type $a=0, b=-z, c=\omega z, d=\omega^2z$. – Jyrki Lahtonen Dec 06 '23 at 14:18
@JyrkiLahtonen I see. I guess we should specify that $a\neq0$. Otherwise, we have $$z^k+(\omega z)^k+ (\omega^2 z)^k = 0$$ for any odd exponent $k\neq3m$. – Tito Piezas III Dec 06 '23 at 14:33
@TitoPiezasIII you seem to like number theory. Im not sure if it is ok to ask like this but, maybe this is something you can answer : Or maybe even correct or improve if needed.
https://math.stackexchange.com/questions/4815649/prime-twin-counting-by-pi-2t2-sum-2jt2-2-omegaj-1-2-lf
– mick Dec 06 '23 at 19:36
@Tito Piezas III, For degree 13 if we consider the simillar form as in degree 11, (1+p)^13+(1-p)^13=(1+q)^13+(1-q)^13, then we have the equation, [(m^6-n^6)+22(m^5-n^5)+99(m^4-n^4)+132(m^3-n^3)+55(m^2-n^2)+6(m-n)]=0. where (m,n)=(p^2,q^2). Hence we need to solve a sixth degree equation (in disquise) for for imaginary numbers. – David Jan 13 '24 at 19:15

The diophantine $a^{11} + b^{11} = c^{11} + d^{11} = s$ for Eisenstein integers.

1 Answers1

Linked