2 Proof of Theorem 1.1

Let $\mathcal {O} = \mathbb {Z}[\sqrt d]$ , let $\sigma $ denote the nontrivial element of $\mathrm {Gal}(K / \mathbb {Q})$ and define the regular representation

$$ \begin{align*} \phi: \mathcal{O} \to M_2(\mathbb{Z}), \quad \phi(a + b \sqrt d) = \bigg(\begin{matrix} a & b \\ d b & a \end{matrix}\bigg), \end{align*} $$

which encodes multiplication on $\mathcal {O}$ in the basis $(1, \sqrt d)$ . Then, $\phi $ is an injective ring homomorphism with $\det \phi (\alpha ) = N(\alpha ) := \alpha \sigma (\alpha )$ , and $\phi $ restricts to a bijection between $\{\alpha \in \mathcal {O} : N(\alpha ) = l\}$ and $G_2(d, l)$ .

Proof of part (i).

Suppose $(X, Y, Z) \in G_2(d, l)^3$ is a solution of (1.1). Write ${X = \phi (\alpha )}$ , $Y = \phi (\beta )$ , $Z = \phi (\gamma )$ with $N(\alpha ) = N(\beta ) = N(\gamma ) = l$ . Applying $\phi ^{-1}$ ,

(2.1) $$ \begin{align} \alpha^n + \beta^n = 2 \gamma^n. \end{align} $$

If $l = 0$ , then $a^2 = d b^2$ with $d$ squarefree forces $a = b = 0$ , so $G_2(d, 0) = \{0\}$ and the only solution is the zero solution, which is trivial. Assume henceforth $l \ne 0$ , so $\alpha , \beta , \gamma $ are nonzero.

Apply $\sigma $ to (2.1) and multiply the resulting equation and (2.1):

$$ \begin{align*} (\alpha^n + \beta^n)(\sigma\alpha^n + \sigma\beta^n) = 4 (\gamma \,\sigma\gamma)^n = 4 l^n. \end{align*} $$

Expanding and using $\alpha \,\sigma \alpha = \beta \,\sigma \beta = l$ ,

$$ \begin{align*} 2 l^n + (\alpha \, \sigma\beta)^n + (\beta \, \sigma\alpha)^n = 4 l^n. \end{align*} $$

Set $u := \alpha \, \sigma \beta $ . Then, $\beta \, \sigma \alpha = \sigma (u)$ and $u \, \sigma (u) = N(\alpha ) N(\beta ) = l^2$ , so $\sigma (u) = l^2 / u$ . Substituting,

$$ \begin{align*} u^n + (l^2 / u)^n = 2 l^n, \quad \text{equivalently}, \quad (u^n - l^n)^2 = 0. \end{align*} $$

Hence, $u^n = l^n$ . Dividing by $(\alpha \, \sigma \alpha )^n = l^n$ and applying $\sigma $ ,

$$ \begin{align*} (\beta / \alpha)^n = 1. \end{align*} $$

Therefore, $\zeta := \beta / \alpha $ lies in $K \cap \mu _n = \mu _{\gcd (n, w_K)}$ , the largest subgroup of $\mu _n$ contained in $K$ . Substituting $\beta = \zeta \alpha $ into (2.1) and using $\zeta ^n = 1$ gives $2 \alpha ^n = 2 \gamma ^n$ , whence $\xi := \gamma / \alpha $ also lies in $\mu _{\gcd (n, w_K)}$ .

The solution is trivial precisely when $(\zeta , \xi ) \in \{\pm 1\}^2$ . If $\gcd (n, w_K) \leqslant 2$ , then $\mu _{\gcd (n, w_K)} \subseteq \{\pm 1\}$ and every solution is trivial. This proves part (i).

3 Solution counts

The proof of Theorem 1.1 extracts more than nontriviality; it yields a complete description of the solution set as a fibre product over $\mathcal {O}$ .

Proposition 3.1. Fix $d$ a nonzero squarefree integer, $l \in \mathbb {Z} \setminus \{0\}$ and $n \geqslant 2$ . Set $\mathcal {O} =~\mathbb {Z}[\sqrt d]$ . The map $(\alpha , \zeta , \xi ) \mapsto (\phi (\alpha ), \phi (\zeta \alpha ), \phi (\xi \alpha ))$ is a bijection between

$$ \begin{align*} \{ (\alpha, \zeta, \xi) : \alpha \in \mathcal{O},\ N(\alpha) = l,\ \zeta, \xi \in \mu_n \cap K,\ \zeta \alpha \in \mathcal{O},\ \xi \alpha \in \mathcal{O} \} \end{align*} $$

and the set of ordered solutions of (1.1) in $G_2(d, l)^3$ .

For $d \notin \{-1, -3\}$ , $\mu _n \cap K = \{\pm 1\}$ and the bijection collapses to the trivial set $\mathcal {T}$ . The nontrivial cases yield exact counts. For a positive integer $l$ , let

$$ \begin{align*}r_d(l) := \#\{(a, b) \in \mathbb{Z}^2 : a^2 - d b^2 = l\}.\end{align*} $$

The asymmetry in trivial counts between odd and even $n$ is because, for $n$ odd and $\alpha \ne 0$ , the equation $\alpha ^n + (-\alpha )^n = 2 \gamma ^n$ reduces to $0 = 2 \gamma ^n$ , forcing $\gamma = 0$ and hence $l = 0$ . Thus, only $(\zeta , \xi ) = (1, 1)$ yields a triple in $\mathcal {T}$ when $n$ is odd.

4 Proof of Theorem 1.2

Let $K$ be a totally real number field of degree $k$ , with embeddings $\sigma _1, \ldots , \sigma _k$ mapping $K \hookrightarrow \mathbb {R}$ . Let $\mathcal {O} \subseteq K$ be an order, $B = (e_1, \ldots , e_k)$ a $\mathbb {Z}$ -basis and $\phi : \mathcal {O} \to ~M_k(\mathbb {Z})$ the regular representation in basis $B$ . As in the quadratic case, $\phi $ is an injective ring homomorphism with $\det \phi (\alpha ) = N_{K/\mathbb {Q}}(\alpha ) = \prod _{i=1}^k \sigma _i(\alpha )$ , and $\phi $ restricts to a bijection between $\{\alpha \in \mathcal {O} : N(\alpha ) = l\}$ and $G_\phi (\mathcal {O}, l)$ .

Suppose $(X, Y, Z) = (\phi (\alpha ), \phi (\beta ), \phi (\gamma ))$ is a solution of (1.1) in $G_\phi (\mathcal {O}, l)^3$ with $l \ne ~0$ . Then, $\alpha , \beta , \gamma $ are nonzero elements of $\mathcal {O}$ with $\alpha ^n + \beta ^n = 2 \gamma ^n$ and $N(\alpha ) = N(\beta ) = N(\gamma ) =~l$ .

Take the product of the conjugate equations $\sigma _i(\alpha )^n + \sigma _i(\beta )^n = 2 \sigma _i(\gamma )^n$ over $i =1, \ldots , k$ :

$$ \begin{align*} \prod_{i = 1}^{k} ( \sigma_i(\alpha)^n + \sigma_i(\beta)^n ) = 2^k \prod_{i = 1}^{k} \sigma_i(\gamma)^n = 2^k N(\gamma)^n = 2^k l^n. \end{align*} $$

Factor out $\prod _i \sigma _i(\beta )^n = N(\beta )^n = l^n$ on the left and set $\zeta := \alpha / \beta \in K^\times $ , so that $\sigma _i(\zeta ) =~\sigma _i(\alpha ) / \sigma _i(\beta )$ . Then,

(4.1) $$ \begin{align} l^n \prod_{i=1}^{k} ( 1 + \sigma_i(\zeta)^n ) = 2^k l^n \quad \implies \quad \prod_{i=1}^{k} ( 1 + \sigma_i(\zeta)^n ) = 2^k. \end{align} $$

Furthermore, $N(\zeta ) = N(\alpha ) / N(\beta ) = 1$ , so

(4.2) $$ \begin{align} \prod_{i=1}^{k} \sigma_i(\zeta)^n = N(\zeta)^n = 1. \end{align} $$

Set $t_i := \sigma _i(\zeta )^n$ . Since $K$ is totally real, $\sigma _i(\zeta ) \in \mathbb {R}^\times $ . Since $n$ is even, $t_i = \sigma _i(\zeta )^n> 0$ . By the AM-GM inequality, $1 + t_i \geqslant 2 \sqrt {t_i}$ for every $i$ , with equality if and only if $t_i = 1$ . Taking the product over $i$ and using (4.2),

$$ \begin{align*} \prod_{i=1}^{k} (1 + t_i) \geqslant 2^k \bigg( \prod_{i=1}^{k} t_i \bigg)^{1/2} = 2^k. \end{align*} $$

Equality in (4.1) forces equality in each AM-GM step; hence, $t_i = 1$ for every $i$ , that is, $\sigma _i(\zeta )^n = 1$ for every embedding $\sigma _i$ .

Since $\sigma _i(\zeta )^n = 1$ for every embedding $\sigma _i$ and the $\sigma _i$ are injective, $\zeta ^n = 1$ . Since $\zeta \in K$ and $\sigma _i(\zeta ) \in \mathbb {R}$ with $\sigma _i(\zeta )^n = 1$ and $n$ even, $\sigma _i(\zeta ) \in \{\pm 1\}$ . As $K$ is totally real, $\mu (K) \subseteq \{\pm 1\}$ , so $\zeta = \pm 1$ , that is, $\alpha = \pm \beta $ .

Substituting $\beta = \pm \alpha $ into $\alpha ^n + \beta ^n = 2 \gamma ^n$ and using $n$ even gives $2 \alpha ^n = 2 \gamma ^n$ ; hence, $(\gamma / \alpha )^n = 1$ and by the same argument, $\gamma = \pm \alpha $ . Therefore, $(X, Y, Z) = (\phi (\alpha ), \pm \phi (\alpha ), \pm \phi (\alpha ))$ , which lies in the trivial set $\mathcal {T}$ .

For $k = 1$ , we have $\mathcal {O} = \mathbb {Z}$ and $G_\phi (\mathbb {Z}, l) = \{l\}$ is a singleton: the only element is $\phi (l) = l$ itself, giving the unique triple is $(l, l, l)$ , which is trivially a solution. No nontrivial arithmetic content arises in this case; the theorem is vacuous for $k = 1$ and its substance lies entirely in $k \geqslant 2$ . In particular, Theorem 1.2 should not be read as asserting anything about the unconstrained integer equation $x^n + y^n = 2z^n$ , which for $n = 2$ has infinitely many nontrivial solutions. Indeed, $x^2 + y^2 = 2z^2$ has infinitely many integer solutions with $y = 1$ , since these come from the Pell equation $x^2 - 2z^2 = -1$ , whose solutions $(x, z) = (1, 1), (7, 5), (41, 29), \ldots $ are infinite in number (giving $7^2 + 1^2 = 2 \cdot 5^2$ , $41^2 + 1^2 = 2 \cdot 29^2$ , and so on).

5 Further research

The arguments above settle (1.1) completely in the matrix classes of quadratic fields and in those of totally real fields for even $n$ , leaving the following questions natural and, to our knowledge, open.

After the reduction of Section 4, Question 5.1 is equivalent to asking whether a unit $\zeta \in \mathcal {O}_K^\times $ with $N(\zeta ) = 1$ , $\zeta \ne 1$ , can satisfy

(5.1) $$ \begin{align} \operatorname{\mathrm{Tr}}_{K / \mathbb{Q}}(\zeta^n) + \operatorname{\mathrm{Tr}}_{K / \mathbb{Q}}(\zeta^{-n}) = 2k. \end{align} $$

The following partial result narrows the search significantly.

Proposition 5.2. Let $K$ be a totally real number field of degree $k$ , $n \geqslant 1$ an integer, and $\zeta \in K^\times $ a totally positive element with $N_{K/\mathbb {Q}}(\zeta ) = 1$ and $\zeta \ne 1$ . Then,

$$ \begin{align*} \operatorname{\mathrm{Tr}}_{K / \mathbb{Q}}(\zeta^n) + \operatorname{\mathrm{Tr}}_{K / \mathbb{Q}}(\zeta^{-n})> 2 k. \end{align*} $$

In particular, no totally positive unit other than $\zeta = 1$ satisfies (5.1).

Proof. For each real embedding $\sigma _i$ , $\sigma _i(\zeta )> 0$ and the AM-GM inequality gives $\sigma _i(\zeta )^n + \sigma _i(\zeta )^{-n} \geqslant 2$ , with equality if and only if $\sigma _i(\zeta ) = 1$ . Summing over $i$ ,

$$ \begin{align*} \operatorname{\mathrm{Tr}}(\zeta^n) + \operatorname{\mathrm{Tr}}(\zeta^{-n}) = \sum_{i=1}^{k} ( \sigma_i(\zeta)^n + \sigma_i(\zeta)^{-n} ) \geqslant 2 k. \end{align*} $$

Equality holds if and only if $\sigma _i(\zeta ) = 1$ for every $i$ , that is, $\zeta = 1$ . As $\zeta \ne 1$ , the inequality is strict.

Any solution of (5.1) must therefore have $\sigma _i(\zeta ) < 0$ for at least one embedding $\sigma _i$ . Since $N(\zeta ) = \prod _i \sigma _i(\zeta ) = 1> 0$ , the number of negative embeddings is even and at least two. For totally real cubic fields, this leaves exactly one possibility.

We have verified computationally that for the fundamental units $\varepsilon _K$ and their powers $\varepsilon _K^{\,j}$ for $1 \leqslant j \leqslant 50$ in the totally real cubic fields defined by $x^3 - 3 x - 1$ (discriminant $81$ , the maximal real subfield of $\mathbb {Q}(\zeta _9)$ ) and $x^3 + x^2 - 2 x - 1$ (discriminant $49$ , the maximal real subfield of $\mathbb {Q}(\zeta _7)$ ), and for one norm- $1$ unit per field across all irreducible totally real cubic polynomials of the form $x^3 + b x^2 + c x - 1$ with $|b|, |c| \leqslant 6$ , no $\zeta $ satisfies (5.1) for odd $n \in [3, 15]$ . We caution that a totally real cubic field has unit rank $2$ by Dirichlet’s theorem, so the norm- $1$ unit group is infinite; our search covers units of small height but is not exhaustive. Full details of these computations are available from the authors on request. The values of $\operatorname {\mathrm {Tr}}(\zeta ^n) + \operatorname {\mathrm {Tr}}(\zeta ^{-n})$ in these examples grow rapidly in absolute value with $n$ , since at least one embedding has $|\sigma _i(\zeta )|> 1$ and dominates. We conjecture that (5.1) has no solutions in any totally real number field of degree $\geqslant 3$ for odd $n$ .

The analogue of (4.1) for CM fields involves complex moduli, and the AM-GM step does not transfer directly because $\sigma _i(\zeta ) \in \mathbb {C} \setminus \mathbb {R}$ in general. The triangle inequality $|1 + \sigma _i(\zeta )^n| \leqslant 2$ gives $\prod _i |1 + \sigma _i(\zeta )^n| \leqslant 2^k$ with equality only when each $\sigma _i(\zeta )^n = 1$ , but $\prod _i (1 + \sigma _i(\zeta )^n) = 2^k$ is a stronger condition that we do not see how to exploit without additional input.

This is the natural framework in which the exceptional pairs $(d, n) = (-1, 4 \cdot \mathbb {N})$ and $(-3, 3 \cdot \mathbb {N})$ of Theorem 1.1 sit: these are precisely the embeddings of $\mathbb {Q}(\sqrt d)$ as subfields of $\mathbb {Q}(\zeta _4)$ and $\mathbb {Q}(\zeta _6)$ , respectively. We expect a clean answer in terms of the inclusion lattice of subfields of $\mathbb {Q}(\zeta _m)$ .