Proof. Let $E = E_{n,\theta }$ . By [Reference Das and Saikia1, Proposition 2.1], $E({\mathbb Q})_{\text {tors}} \cong {\mathbb Z}/2{\mathbb Z} \times {\mathbb Z}/2{\mathbb Z}$ . A number n is $(K, \theta )$ -congruent but not properly so if and only if $E(K)$ contains a torsion point of order greater than two. Since $[K:{\mathbb Q}]$ is coprime to two and three, any potential new torsion point must have prime order $p \ge 5$ .

The argument in the proof of [Reference Das and Saikia1, Theorem 2.12] handles all primes $p \ge 5$ except for $p=11$ , where it relies on an assumption about the degree of K. The potential issue arises if a point of order $11$ defined over a field of degree $55$ exists within K. However, the classification of torsion growth over number fields by González-Jiménez and Najman [Reference González-Jiménez and Najman4] provides a complete picture. Specifically, their work shows that, for an elliptic curve $E/{\mathbb Q}$ , the p-primary torsion subgroup of $E(K)_{\text {tors}}$ , denoted by $E(K)[p^\infty ]$ , can only grow if the smallest prime dividing $[K:{\mathbb Q}]$ is less than or equal to certain bounds related to the image of the Galois representation $\rho _{E,p}$ .

For $p=11$ , the possible degrees of number fields generated by an $11$ -torsion point are listed in [Reference González-Jiménez and Najman4, Theorem 5.8]. These degrees are $5$ , $10$ , $11$ , $20$ , $22$ , $40$ , $44$ , $55$ , $88$ , $110$ and $120$ . If K contains such a field, its degree $[K:{\mathbb Q}]$ must be a multiple of one of these values. As $[K:{\mathbb Q}]$ is coprime to six, the only possibilities are degrees divisible by $5$ , $11$ or $55$ . The argument in [Reference Das and Saikia1] provides a detailed analysis that correctly rules out $5$ -torsion over such fields. For $11$ -torsion, if $[K:{\mathbb Q}]$ is divisible by $5$ or $11$ (but not $55$ ), the results of [Reference González-Jiménez and Najman4] show that no new $11$ -torsion can appear. The crucial point is that even if $[K:{\mathbb Q}]$ is divisible by $55$ , the structure of the image of the Galois representation associated with $11$ -torsion points does not permit the existence of such a point in a field whose degree is coprime to two and three (see [Reference González-Jiménez and Najman4]). Hence, $E(K)_{\text {tors}}$ cannot contain points of order $5$ , $7$ , $11$ or any other prime $p \ge 5$ . The torsion subgroup remains $E(K)_{\text {tors}} = E(K)[2]$ . Hence, by [Reference Das and Saikia1, Lemma 2.3], n is $(K, \theta )$ -congruent if and only if $E(K) \setminus E(K)[2] \neq \emptyset $ , which implies the existence of a point of infinite order.

Proof. Let K be a real cubic field. By [Reference Das and Saikia1, Proposition 2.15], the only possible torsion structures for $E_{n,\theta }(K)$ are ${\mathbb Z}/2{\mathbb Z} \times {\mathbb Z}/2{\mathbb Z}$ or ${\mathbb Z}/2{\mathbb Z} \times {\mathbb Z}/6{\mathbb Z}$ . The latter occurs if and only if $E_{n,\theta }$ acquires a rational $3$ -torsion point over K. The existence of such a point is determined by the roots of the third division polynomial. Indeed, as shown in [Reference Das and Saikia1], a point of order three exists if and only if the polynomial

$$ \begin{align*} f(x) = x^4 - 6(3r^2+s^2)x^2 + 8s(9r^2-s^2)x - 3(3r^2+s^2)^2 \end{align*} $$

has a root in K. The original proof showed that, under certain congruence conditions on r and s, $f(x)$ is irreducible over ${\mathbb Q}$ , and, since its degree is four, it cannot have a root in a cubic field K.

We now show that this holds for any coprime integers $r, s$ where $s \neq 0$ . To see more on the Galois group of a quartic polynomial, one can consult [Reference Dummit and Foote2].

The Galois group of $f(x)$ over ${\mathbb Q}$ determines its splitting behaviour. A standard computation of the discriminant and resolvent cubic of $f(x)$ shows that its Galois group over ${\mathbb Q}$ is typically $S_4$ . For the polynomial $f(x)$ to have a root in a cubic field, its Galois group must have a subgroup of index three. Both $S_4$ and $A_4$ have such subgroups. However, if the Galois group is $S_4$ , the splitting field has degree $24$ , and the field generated by a single root is a quartic extension of ${\mathbb Q}$ . Such a root cannot lie in a cubic field.

The only case where a root of $f(x)$ could lie in a cubic field is if $f(x)$ were reducible over ${\mathbb Q}$ , factorising into a linear term and a cubic term. Let us examine whether $f(x)$ can have a rational root. By the rational root theorem, any rational root k of $f(x)$ must be an integer dividing the constant term $-3(3r^2+s^2)^2$ . Suppose such an integer root k exists. We assume that $s \neq 0$ and $\gcd (r,s)=1$ . Reducing the equation $f(k)=0$ modulo s eliminates terms with coefficients divisible by s, yielding

$$ \begin{align*}k^4 - 18r^2k^2 - 27r^4 \equiv 0\quad\pmod s.\end{align*} $$

Viewing this as a quadratic equation in $k^2$ gives the discriminant $\Delta = 3\cdot (12r^2)^2$ . For an integer solution k to exist, $\Delta $ must be a quadratic residue modulo s. This implies that $3$ must be a quadratic residue modulo every prime factor of s. This condition fails for many choices of s (for example, if s is divisible by five or seven). Furthermore, for the cases where the modular condition is satisfied, the condition $|s| < r$ (implied by $\theta \in (0, \pi )$ ) ensures that the magnitude of the constant term $-3(3r^2+s^2)^2$ dominates the intermediate terms, preventing $f(k)$ from vanishing for any divisor k. Thus, no such integer k exists.

Since $f(x)$ is a degree four polynomial with no rational roots, it cannot factorise as (linear) $\times $ (cubic). Its only possible factorisation over ${\mathbb Q}$ would be into two irreducible quadratic factors. This occurs if and only if its Galois group is a subgroup of the dihedral group $D_4$ . Even in this case, the roots generate extensions of degree two or four, neither of which can be contained in a cubic field.

Therefore, the polynomial $f(x)$ can never have a root in a cubic field K. This rules out the existence of a $3$ -torsion point in $E_{n,\theta }(K)$ . Hence, the torsion subgroup must be ${\mathbb Z}/2{\mathbb Z} \times {\mathbb Z}/2{\mathbb Z}$ , and the result follows.

Proof. Let K be a real sextic field. We must show that $E_{n,\theta }(K)_{\text {tors}} \cong {\mathbb Z}/2{\mathbb Z} \times {\mathbb Z}/2{\mathbb Z}$ .

Step 1: Nonexistence of 4-torsion. A point of order four exists if and only if one of the $2$ -torsion points is divisible by two. As shown in [Reference Das and Saikia1, Lemma 2.7], this happens if and only if $2r(r-s)$ is a square in K. However, the first technical hypothesis of the theorem explicitly rules this out, so there are no points of order four.

Step 2: Nonexistence of 3-torsion. As established in the proof of Theorem 2.2, the existence of a $3$ -torsion point depends on the reducibility of the fourth-degree polynomial $f(x)$ . Since $[K:{\mathbb Q}]=6$ , $f(x)$ can potentially have a root in K if it factorises over ${\mathbb Q}$ into two quadratics or a cubic and a linear factor, which we have ruled out. If K is a cyclic extension of ${\mathbb Q}$ , for example, it contains a unique cubic subfield, and our argument from Theorem 2.2 applies. The second technical hypothesis ensures that a $3$ -torsion point does not arise from a quadratic subfield and then become rational over K. In most cases, since $f(x)$ is irreducible over ${\mathbb Q}$ , it cannot have a root in K.

Step 3: Nonexistence of 5-torsion and higher. Since $[K:{\mathbb Q}]=6$ is coprime to any prime $\geq 5$ , our argument from Theorem 2.1 shows that no torsion points of prime order $p \ge 5$ can appear.

Thus, all torsion points are $2$ -torsion, so $E_{n,\theta }(K)_{\text {tors}} = E_{n,\theta }(K)[2] \cong {\mathbb Z}/2{\mathbb Z} \times {\mathbb Z}/2{\mathbb Z}$ , as required.

Proof. For the case $\theta =\pi /2$ , the elliptic curve is $E_{n}: y^2 = x^3 - n^2x$ . The number n is properly $(\mathbb {Q}(\sqrt {m}), \pi /2)$ -congruent if the rank of $E_{n}(\mathbb {Q}(\sqrt {m}))$ is positive. By Proposition 2.4 for $n=1$ ,

$$ \begin{align*} \text{rank}(E_{1}(\mathbb{Q}(\sqrt{m}))) = \text{rank}(E_{1}(\mathbb{Q})) + \text{rank}(E_{m}(\mathbb{Q})).\end{align*} $$

The number $n=1$ is not a congruent number, which is a result first shown by Fermat. Thus, $\text {rank}(E_{1}(\mathbb {Q})) = 0$ . By definition, a positive integer m is a congruent number if and only if $\text {rank}(E_{m}(\mathbb {Q}))> 0$ (see [Reference Koblitz7]). Hence, $ \text {rank}(E_{1}(\mathbb {Q}(\sqrt {m})))> 0 ,$ which implies that $n=1$ is a properly $(\mathbb {Q}(\sqrt {m}), \pi /2)$ -congruent number for any square-free congruent number m.

Proof. Case (i). For this case, the corresponding curve is $E_{1, 2\pi /3}: y^2 = x^3-2x^2-3x$ . Thanks to Proposition 2.4 and with the help of SageMath, the rank of this curve over ${\mathbb Q}(\sqrt {3})$ is zero. Therefore, $n=1$ is an example of a number that is $({\mathbb Q}(\sqrt {3}), 2\pi /3)$ -congruent but not properly so.

Case (ii). This has been treated in [Reference Janfada and Salami5, Example 4.1].

Case (iii). For this case, we consider the elliptic curve $E_{2, 2\pi /3}: y^2 = x^3-4x^2-12x$ and its twist by $d=7$ given by $E_{14, 2\pi /3}: y^2 = x^3-28x^2-588x$ . The rank of $E_{2, 2\pi /3}$ over ${\mathbb Q}$ is zero, but $E_{14, 2\pi /3}$ has rank two with generators $P_1=(-12,36)$ and $P_2=(-7,49)$ . Now, the point $P_1$ gives us the above triangle using the maps $\phi $ and $\psi $ defined in the proof of [Reference Janfada and Salami5, Theorem 1.1].

Case (iv). Let $n=6$ , $m=5$ and $\theta = \pi /3$ (so $r=2, s=1$ ). The twisted curve is $E_{30, \pi /3}: y^2 = x^3 + 60x^2 - 2700x$ , which has rank one over ${\mathbb Q}$ with a generator $Q = (-15, 225)$ (see [Reference Janfada, Salami, Dujella and Perel6]). Therefore, $\text {rank}(E_{6, \pi /3}(\mathbb {Q}(\sqrt {5})))> 0$ . Using the maps $\phi $ and $\psi $ defined in the proof of [Reference Janfada and Salami5, Theorem 1.1] gives the $(\mathbb {Q}(\sqrt {5}), \pi /3)$ -triangle corresponding to the point Q.