POINTS ON $x^4+y^4=z^4$ OVER QUADRATIC EXTENSIONS OF ${\mathbb {Q}}(\zeta _8)(T_1,T_2,\ldots ,T_n)$

Abstract

Ishitsuka et al. [‘Explicit calculation of the mod 4 Galois representation associated with the Fermat quartic’, Int. J. Number Theory 16(4) (2020), 881–905] found all points on the Fermat quartic ${F_4\colon x^4+y^4=z^4}$ over quadratic extensions of ${\mathbb {Q}}(\zeta _8)$, where $\zeta _8$ is the eighth primitive root of unity $e^{i\pi /4}$. Using Mordell’s technique, we give an alternative proof for the result of Ishitsuka et al. and extend it to the rational function field ${\mathbb {Q}}({\zeta _8})(T_1,T_2,\ldots ,T_n)$.

1 Introduction

The problem of finding points on the Fermat quartic

$$ \begin{align*} F_4\colon x^4+y^4=z^4 \end{align*} $$

over number fields has been studied by several authors. Fermat showed that $F_4$ only has trivial points over the rational numbers, where a trivial point on $F_4$ is a point $[x:y:z]$ with $xyz=0$ . Aigner [1] showed that if $F_4$ has nontrivial points in a quadratic number field ${\mathbb {Q}}(\sqrt {d})$ , then $d=-7$ . Faddeev [5] later found all points on $F_4$ over all quadratic number fields and all cubic number fields. Bremner and Choudhry [3] showed that $F_4$ only has trivial points in any cyclic cubic number field. Recently, Ishitsuka et al. found all points on $F_4$ over quadratic extensions of ${\mathbb {Q}}(\zeta _8)$ .

Theorem 1.1 (Ishitsuka et al. [6, Theorem 7.3]).

There are $188$ points on $F_4$ defined over quadratic extensions of ${\mathbb {Q}}(\zeta _8)$ :

1. (A) $12$ trivial points defined over ${\mathbb {Q}}(\zeta _8)$ : $[1:\zeta _8^{j}:0], [0:\pm 1:1], [0:\pm {\zeta _8}^2:1]$ , and $[\pm 1:0:1], [\pm {\zeta _8}^2:0:1 ], j=1,3,5,7$ ;

2. (B) $48$ points defined over ${\mathbb {Q}}(2^{1/4}\zeta _8)$ : $[2^{1/4}\zeta _8^{2i}:\zeta _8^{1+2j}:1], [\zeta _8^{1+2j}:2^{1/4}\zeta _8^{2i}:1]$ , $[2^{-1/4}\zeta _8^{2i}:2^{-1/4}\zeta _8^{2j}:1], 0\leq i,j\leq 3$ ;

3. (C) $32$ points defined over ${\mathbb {Q}}(\zeta _3,\zeta _8)$ : $[\zeta _3\zeta _8^{1+2i}:\zeta _3^{2}\zeta _8^{1+2j}:1], [\zeta _3^2\zeta _8^{1+2i}:\zeta _3\zeta _8^{1+2j}:1]$ , $0\leq i,j\leq 3$ ;

4. (D) $96$ points defined over ${\mathbb {Q}}(\sqrt {-7},\zeta _8)$ : $[\alpha \zeta _8^{2i}:\overline {\alpha }\zeta _8^{2j}:1], [\overline {\alpha }\zeta _8^{2i}:\alpha \zeta _8^{2j}:1]$ , $[\overline {\alpha }\zeta _8^{7+2j}:\zeta _8^6:\alpha \zeta _8^{2i}],\ \ [\alpha \zeta _8^{7+2j}:\zeta _8^6:\overline {\alpha }\zeta _8^{2i}],\ \ [1\kern1.9pt{:}\kern1.9pt\alpha \zeta _8^{1+2i}\kern1.9pt{:}\kern1.9pt\overline {\alpha }\zeta _8^{2+2j}]$ , $[1\kern1.9pt{:}\kern1.9pt\overline {\alpha }\zeta _8^{1+2i}\kern1.9pt{:}\kern1.9pt\alpha \zeta _8^{2+2j}], 0\kern2pt{\leq}\kern2pt i,j\kern2pt{\leq}\kern2pt 3$ ,

where $\zeta _3=e^{2i\pi /3}$ , $\alpha =(1+\sqrt {-7})/2$ and $\overline {\alpha }=(1-\sqrt {-7})/2$ .

In this paper, we give an alternative proof for Theorem 1.1 and its extension to rational function fields.

Theorem 1.2. There are $188$ points on $F_4$ defined over quadratic extensions of ${\mathbb {Q}}(\zeta _8)(T_1,T_2,\ldots ,T_n)$ . These points are the $188$ points in Theorem 1.1.

Before moving on to the proof of Theorem 1.2, we note that our approach is completely different from the approach of Ishitsuka et al. In [6], the authors use techniques from Galois representation theory and their proof relies on Rohrlich’s result [11, Corollary 1, page 117] and Faddeev’s result [5, Section 3, page 1150]. This paper is modelled on Mordell’s paper [9], where he reproved Faddeev’s result [5], see also [10, Theorem 4, pages 116–118]. The advantage of Mordell’s approach is that it is simple, easy to use and concrete in calculations. For some other applications of this approach, see Li [7] and Manley [8].

2 Some preliminary results

Lemma 2.1. Let k be a field of characteristic not $2$ . Let $K=k(T_1,T_2,\ldots ,T_n)$ be the function field over k generated by n algebraically independent variables $T_1,T_2,\ldots ,T_n$ . Let E be the elliptic curve over K given by $y^2=x^3+Ax+B$ , where $A,B\in k$ . Then, $E(K)=E(k)$ .

Proof. The following proof of Lemma 2.1 is due to Professor Andrew Bremner (personal communication). We use induction on n. When $n=1$ , see Cohen [4, Proposition 7.3.2, pages 487–488]. Assume that Lemma 2.1 is true for n. Let ${K=k(T_1,T_2,\ldots ,T_n)}$ . Consider the elliptic curve E over $L=k(T_1,T_2,\ldots ,T_{n+1})$ given by $y^2=x^3+Ax+B$ , where $A,B\in k$ . Since $L=K(T_{n+1})$ and $k\subset K$ , by induction,

$$ \begin{align*}E(L)=E(K(T_{n+1}))=E(K)=E(k).\end{align*} $$

The proof is complete.

Lemma 2.1 enables us to verify Lemmas 2.2, 2.3 and 2.4 using MAGMA [2].

Lemma 2.2. All ${\mathbb {Q}}(\zeta _8)(T_1,T_2,\ldots ,T_n)$ -points on $y^2=2x(x^2+1)$ are

$$ \begin{align*}\infty,\, (0,0),\, (1,\pm 2),\, (\pm {\zeta_8}^2, 0),\, (-1, \pm 2{\zeta_8}^2),\end{align*} $$

$$ \begin{align*}({\zeta_8}^3+{\zeta_8}^2+{\zeta_8},\pm (2{\zeta_8}^3+2{\zeta_8}^2-2)),\, ({\zeta_8}^3-{\zeta_8}^2+{\zeta_8},\pm (2{\zeta_8}^2-2{\zeta_8}+2)),\end{align*} $$

$$ \begin{align*}(-({\zeta_8}^3+{\zeta_8}^2+{\zeta_8}),\pm (2{\zeta_8}^2+2{\zeta_8}+2)),\,(-({\zeta_8}^3-{\zeta_8}^2+{\zeta_8})),\pm (2{\zeta_8}^3-2{\zeta_8}^2+2)). \end{align*} $$

Lemma 2.3. All ${\mathbb {Q}}(\zeta _8)(T_1,T_2,\ldots ,T_n)$ -points on $y^2=2x(x^2-1)$ are

$$ \begin{align*} \infty,\, (0,0),\, (\pm 1,0),\, ({\zeta_8}^2,\pm 2{\zeta_8}^3),\, (-{\zeta_8}^2,\pm 2{\zeta_8}),\end{align*} $$

$$ \begin{align*}({\zeta_8}^3-{\zeta_8}-1,\pm 2({\zeta_8}^3+2{\zeta_8}^2+2{\zeta_8})),\, ({\zeta_8}^3-{\zeta_8}+1,\pm (2{\zeta_8}^3-2{\zeta_8}+2)),\end{align*} $$

$$ \begin{align*}(-{\zeta_8}^3+{\zeta_8}+1,\pm (2{\zeta_8}^3-2{\zeta_8}-2)),\, (-{\zeta_8}^3+{\zeta_8}-1,\pm (2{\zeta_8}^3-2{\zeta_8}^2+2{\zeta_8})).\end{align*} $$

Lemma 2.4. All ${\mathbb {Q}}(\zeta _8)(T_1,T_2,\ldots ,T_n)$ -points on $s^2=t^4-1$ are

$$ \begin{align*}\infty,\,(0 , \pm {\zeta_8}^2),\, (\pm 1, 0),\, (\pm {\zeta_8}^2, 0),\, (\pm {\zeta_8}, \pm ({\zeta_8}^3 + {\zeta_8}) ),\, (\pm {\zeta_8}^3 , \pm ({\zeta_8}^3 + {\zeta_8})). \end{align*} $$

3 Proof of Theorem 1.2

Let $K={\mathbb {Q}}({\zeta _8})(T_1,T_2,\ldots ,T_n)$ . Let $L=K(\sqrt {d})$ be a quadratic extension of K, where $d\in K$ and $\sqrt {d}\not \in K$ . Assume $[x:y:z]\in F_4(L)$ .

If $x=0$ , then $y^4=z^4$ . Hence,

(3.1) $$ \begin{align} [x:y:z]=[0:\pm 1:1], [0:\pm \zeta_8^2:1].\end{align} $$

Similarly, if $y=0$ , then

(3.2) $$ \begin{align} [x:y:z]=[\pm 1: 0:1], [\pm \zeta_8^2:0:1].\end{align} $$

If $z=0$ , then $x^4+y^4=0$ . Hence,

(3.3) $$ \begin{align} [x:y:z]=[{\zeta_8}^j:1:0],\quad j=1,3,5,7.\end{align} $$

Note that the 12 points in (3.1), (3.2) and (3.3) are the 12 points in Theorem 1.1(A).

Assume $xyz\neq 0$ . Let $z=1$ . Then,

(3.4) $$ \begin{align} x^4+y^4=1. \end{align} $$

Since $x\neq 0$ , $y^2\neq \pm 1$ . From (3.4), $(1+y^2)/x^2=x^2/(1-y^2)$ . Let

(3.5) $$ \begin{align} t=\dfrac{1+y^2}{x^2}\,\bigg(=\dfrac{x^2}{1-y^2}\bigg).\end{align} $$

Then, $t\neq 0$ . If $t=\pm \zeta _8^2$ , by (3.5),

$$ \begin{align*}-1=t^2=\dfrac{1+y^2}{x^2}\cdot \dfrac{x^2}{1-y^2}=\dfrac{1+y^2}{1-y^2}.\end{align*} $$

Hence, $1+y^2=y^2-1$ , which is impossible. If $t=\pm 1$ , by (3.5),

$$ \begin{align*}1=t^2=\dfrac{1+y^2}{x^2}\cdot \dfrac{x^2}{1-y^2}=\dfrac{1+y^2}{1-y^2}.\end{align*} $$

Hence, $y=0$ , which is impossible. Therefore,

(3.6) $$ \begin{align} t\not\in \{0, \pm 1, \pm \zeta_8^2\}. \end{align} $$

It follows from (3.4) and (3.5) that

(3.7) $$ \begin{align} x^2=\dfrac{2t}{t^2+1},\quad y^2=\dfrac{t^2-1}{t^2+1},\end{align} $$

Let $X=x(t^2+1)$ and $Y=xy(t^2+1)$ . From (3.7),

(3.8) $$ \begin{align} X^2=2t(t^2+1),\end{align} $$

(3.9) $$ \begin{align} Y^2=2t(t^2-1).\end{align} $$

Case I: $t\in K$ . Let $X=u+v\sqrt {d}$ and $Y=u_1+v_1\sqrt {d}$ , where $u,v,u_1,v_1\in K$ . From (3.8), $X^2,Y^2\in K$ . Since $\sqrt {d}\not \in K$ , $uv=u_1v_1=0$ .

Case I.1: $v=0$ . Then, $X=u\in K$ . Thus, $(t,u)$ is a K-point on (3.8). Therefore, by Lemma 2.2, $t=\pm ({\zeta _8}^3+{\zeta _8}^2+{\zeta _8}),\pm ({\zeta _8}^3-{\zeta _8}^2+{\zeta _8})$ . If $t={\zeta _8}^3+{\zeta _8}^2+{\zeta _8}$ , then

(3.10) $$ \begin{align} [x:y:z]=[\pm {\zeta_8}^3:\pm \sqrt{{\zeta_8}-{\zeta_8}^3}:1].\end{align} $$

If $t=-({\zeta _8}^3+{\zeta _8}^2+{\zeta _8})$ , then

(3.11) $$ \begin{align} [x:y:z]=[\pm {\zeta_8}:\pm \sqrt{{\zeta_8}-{\zeta_8}^3}:1].\end{align} $$

If $t={\zeta _8}^3-{\zeta _8}^2+{\zeta _8}$ , then

(3.12) $$ \begin{align} [x:y:z]=[\pm {\zeta_8}:\pm \sqrt{{\zeta_8}^3-{\zeta_8}}:1 ].\end{align} $$

If $t=-({\zeta _8}^3-{\zeta _8}^2+{\zeta _8})$ , then

(3.13) $$ \begin{align} [x:y:z]=[\pm {\zeta_8}^3:\pm \sqrt{{\zeta_8}^3-{\zeta_8}}:1]. \end{align} $$

Case I.2: $v_1=0$ . Then, $(t,u_1)$ is a K-point on (3.9). Therefore, by Lemma 2.3, ${t=\pm ({\zeta _8}^3-{\zeta _8}-1), \pm ({\zeta _8}^3-{\zeta _8}+1)}$ . If $t={\zeta _8}^3-{\zeta _8}-1$ , then

(3.14) $$ \begin{align} [x:y:z]=\bigg[{\pm} \sqrt{\dfrac{{\zeta_8}^3 - {\zeta_8}}{2}}:\pm \sqrt{\dfrac{{\zeta_8}-{\zeta_8}^3 }{2}}:1\bigg]. \end{align} $$

If $t=-({\zeta _8}^3-{\zeta _8}-1)$ , then

(3.15) $$ \begin{align} [x:y:z]=\bigg[{\pm} \sqrt{\dfrac{{\zeta_8} - {\zeta_8}^3}{2}}:\pm \sqrt{\dfrac{{\zeta_8}-{\zeta_8}^3 }{2}}:1\bigg].\end{align} $$

If $t={\zeta _8}^3-{\zeta _8}+1$ , then

(3.16) $$ \begin{align} [x:y:z]=\bigg[{\pm} \sqrt{\dfrac{{\zeta_8}^3 - {\zeta_8}}{2}}:\pm \sqrt{\dfrac{{\zeta_8}^3-{\zeta_8}}{2}}:1\bigg].\end{align} $$

If $t=-({\zeta _8}^3-{\zeta _8}+1)$ , then

(3.17) $$ \begin{align} [x:y:z]=\bigg[{\pm} \sqrt{\dfrac{{\zeta_8} - {\zeta_8}^3}{2}}:\pm \sqrt{\dfrac{{\zeta_8}^3-{\zeta_8}}{2}}:1\bigg].\end{align} $$

Case I.3: $vv_1\neq 0$ . Since $X^2,Y^2\in K$ and $X,Y \notin K$ , we have $u=u_1=0$ . It follows from (3.8) and (3.9) that $dv^2=2t(t^2+1) \text { and } dv_1^2=2t(t^2-1)$ . Hence,

(3.18) $$ \begin{align} s^2=(t^2+1)(t^2-1),\end{align} $$

where $s=dvv_1/(2t) \in K$ . By Lemma 2.4, $t=\pm \zeta _8,\pm \zeta _8^3$ . If $t={\zeta _8}^3$ , then

(3.19) $$ \begin{align} [x:y:z]=[\pm \sqrt{{\zeta_8}^3-{\zeta_8}}:\pm {\zeta_8}^3:1]. \end{align} $$

If $t=-{\zeta _8}^3$ , then

(3.20) $$ \begin{align} [x:y:z]=[\pm \sqrt{{\zeta_8}-{\zeta_8}^3}:\pm {\zeta_8}^3:1]. \end{align} $$

If $t={\zeta _8}$ , then

(3.21) $$ \begin{align} [x:y:z]=[\pm \sqrt{{\zeta_8}-{\zeta_8}^3}:\pm {\zeta_8}:1]. \end{align} $$

If $t=-{\zeta _8}$ , then

(3.22) $$ \begin{align} [x:y:z]=[\pm \sqrt{{\zeta_8}^3-{\zeta_8}}:\pm {\zeta_8}:1]. \end{align} $$

Note that the 48 points in (3.10), (3.11), (3.12), (3.13), (3.14), (3.15), (3.16), (3.17), (3.19), (3.20), (3.21) and (3.22) are the 48 points in Theorem (1.1)(B).

Case II: $t\not \in K$ . Let $P(T)\in K[T]$ be the monic minimal polynomial of t over K. Then, $\mathrm {deg}P(T)=2$ .

Step 1: There exist $a,b\in K$ such that $X=at+b.$ By (3.8), the polynomial $2T(T^2+1)-(aT+b)^2$ has a root $T=t$ . Therefore, there exist $c,d\in K$ such that

(3.23) $$ \begin{align} 2T(T^2+1)-(aT+b)^2=P(T)(cT+d). \end{align} $$

Then, $c=2$ and $(-d/c, -ad/c+b)$ is a K-point on (3.8). Hence, by Lemma 2.3,

$$ \begin{align*}-d/c\in \{0, \pm 1, \pm {\zeta_8}^2, \pm ({\zeta_8}^3+{\zeta_8}^2+{\zeta_8}), \pm ({\zeta_8}^3-{\zeta_8}^2+{\zeta_8})\}.\end{align*} $$

Case 1.1: $-d/c=0$ . Then, $b=d=0$ . From (3.23),

(3.24) $$ \begin{align} P(T)=T^2 - \dfrac{a^2}{2}T + 1.\end{align} $$

Case 1.2: $-d/c=1$ . Then, $d=-2$ and $a+b=\pm 2$ . By changing the signs of a and b, we can assume that $a+b=-2$ . From (3.23),

(3.25) $$ \begin{align} P(T)=T^2 + \bigg({-}\dfrac{a^2}{2} + 1\bigg)T + \dfrac{a^2}{2} + 2a + 2.\end{align} $$

Case 1.3: $-d/c=-1$ . Then, $d=2$ and ${a-b=\pm 2 {\zeta _8}^2}$ . We can assume that ${a-b=2{\zeta _8}^2}$ . From (3.23),

(3.26) $$ \begin{align} P(T)=T^2 + \bigg({-}\dfrac{a^2}{2} - 1\bigg)T - \dfrac{a^2}{2} - 2{\zeta_8}^2a + 2.\end{align} $$

Case 1.4: $-d/c={\zeta _8}^2$ . Then, $d=-2{\zeta _8}^2$ and $b=-a{\zeta _8}^2$ . From (3.23),

(3.27) $$ \begin{align} P(T)=T^2 + \bigg({-}\dfrac{a^2}{2} + {\zeta_8}^2\bigg)T + \dfrac{1}{2}{\zeta_8}^2a^2.\end{align} $$

Case 1.5: $-d/c=-{\zeta _8}^2$ . Then, $d=2{\zeta _8}^2$ and $b=a{\zeta _8}^2$ . From (3.23),

(3.28) $$ \begin{align} P(T)=T^2 + \bigg({-}\dfrac{a^2}{2} - {\zeta_8}^2\bigg)T - \dfrac{1}{2}{\zeta_8}^2a^2.\end{align} $$

Case 1.6: $-d/c={\zeta _8}^3+{\zeta _8}^2+{\zeta _8}$ . Then, $d=-2({\zeta _8}^3+{\zeta _8}^2+{\zeta _8})$ . It follows that

$$ \begin{align*} ({\zeta_8}^3+{\zeta_8}^2+{\zeta_8})a+b=\pm 2({\zeta_8}^3+{\zeta_8}^2-1). \end{align*} $$

We can assume that $b=2({\zeta _8}^3+{\zeta _8}^2-1)-({\zeta _8}^3+{\zeta _8}^2+{\zeta _8})a$ . From (3.23),

(3.29) $$ \begin{align} P(T)& =T^2 + \bigg({-}\dfrac{a^2}{2} + ({\zeta_8}^3 + {\zeta_8}^2 + {\zeta_8})\bigg)T + \dfrac{1}{2}({\zeta_8}^3 + {\zeta_8}^2 + {\zeta_8})a^2 \nonumber\\ & \quad + (-2{\zeta_8}^3 - 2{\zeta_8}^2 + 2)a + 2{\zeta_8}^3 - 2{\zeta_8} - 2. \end{align} $$

Case 1.7 $-d/c=-({\zeta _8}^3+{\zeta _8}^2+{\zeta _8})$ . Then, $d=2({\zeta _8}^3+{\zeta _8}^2+{\zeta _8})$ . Then,

$$ \begin{align*}-({\zeta_8}^3+{\zeta_8}^2+{\zeta_8})a+b=\pm 2({\zeta_8}^2+{\zeta_8}+1).\end{align*} $$

We can assume that $b=2({\zeta _8}^2+{\zeta _8}+1)+({\zeta _8}^3+{\zeta _8}^2+{\zeta _8})a$ . From (3.23),

(3.30) $$ \begin{align} P(T)& = T^2 + \bigg({-}\dfrac{a^2}{2} - {\zeta_8}^3 - {\zeta_8}^2 - {\zeta_8}\bigg)T + \dfrac{1}{2}(-{\zeta_8}^3 - {\zeta_8}^2 - {\zeta_8})a^2 \nonumber\\ & \quad + (-2{\zeta_8}^2 - 2{\zeta_8} - 2)a + 2{\zeta_8}^3 - 2{\zeta_8} - 2. \end{align} $$

Case 1.8: $-d/c={\zeta _8}^3-{\zeta _8}^2+{\zeta _8}$ . Then, $d=-2({\zeta _8}^3-{\zeta _8}^2+{\zeta _8})$ and

$$ \begin{align*}({\zeta_8}^3-{\zeta_8}^2+{\zeta_8})a+b=\pm 2({\zeta_8}^2-{\zeta_8}+1).\end{align*} $$

We can assume that

$$ \begin{align*}b=2({\zeta_8}^2-{\zeta_8}+1)-({\zeta_8}^3-{\zeta_8}^2+{\zeta_8})a.\end{align*} $$

From (3.23),

(3.31) $$ \begin{align} P(T)& = T^2 + \bigg({-}\dfrac{a^2}{2} + ({\zeta_8}^3 - {\zeta_8}^2 + {\zeta_8})\bigg)T + \dfrac{1}{2}({\zeta_8}^3 - {\zeta_8}^2 + {\zeta_8})a^2 \nonumber\\ & \quad + (-2{\zeta_8}^2 + 2{\zeta_8} - 2)a - 2{\zeta_8}^3 + 2{\zeta_8} - 2. \end{align} $$

Case 1.9: $-d/c=-({\zeta _8}^3-{\zeta _8}^2+{\zeta _8})$ . Then, $d=2({\zeta _8}^3-{\zeta _8}^2+{\zeta _8})$ and

$$ \begin{align*}-({\zeta_8}^3-{\zeta_8}^2+{\zeta_8})a+b=\pm 2({\zeta_8}^3-{\zeta_8}^2+1).\end{align*} $$

We can assume that $b=2({\zeta _8}^3-{\zeta _8}^2+1)+({\zeta _8}^3-{\zeta _8}^2+{\zeta _8})a$ . From (3.23),

(3.32) $$ \begin{align} P(T)& = T^2 + \bigg({-}\dfrac{a^2}{2} - {\zeta_8}^3 + {\zeta_8}^2 - {\zeta_8}\bigg)T + \dfrac{1}{2}(-{\zeta_8}^3 + {\zeta_8}^2 - {\zeta_8})a^2 \nonumber\\ & \quad + (-2{\zeta_8}^3 + 2{\zeta_8}^2 - 2)a - 2{\zeta_8}^3 + 2{\zeta_8} - 2. \end{align} $$

Step 2: There exist $a_1,b_1\in K$ such that $Y=a_1t+b_1$ . Then, (3.9) shows that the polynomial $2T(T^2-1)-(a_1T+b_1)^2$ has a root $T=t$ . Hence, there exist $c_1,d_1\in K$ such that

(3.33) $$ \begin{align} 2T(T^2-1)-(a_1T+b_1)^2=P(T)(c_1T+d_1).\end{align} $$

Thus, $c_1=2$ and $(-d_1/c_1,-a_1d_1/c_1+b_1)$ is a finite K-point on (3.9). Hence

$$ \begin{align*}-d_1/c_1\in \{0,\pm 1,\pm {\zeta_8}^2, \pm ({\zeta_8}^3-{\zeta_8}-1), \pm ({\zeta_8}^3-{\zeta_8}+1)\}.\end{align*} $$

Case 2.1: $-d_1/c_1=0$ . Then, $b_1=d_1=0$ . From (3.33),

(3.34) $$ \begin{align} P(T)=T^2 - \dfrac{a_1^2}{2}T - 1.\end{align} $$

Case 2.2: $-d_1/c_1=1$ . Then, $d_1=-2$ and $b_1=-a_1$ . From (3.33),

(3.35) $$ \begin{align} P(T)=T^2 + \bigg({-}\dfrac{a_1^2}{2} + 1\bigg)T +\dfrac{a_1^2}{2}.\end{align} $$

Case 2.3: $-d_1/c_1=-1$ . Then, $d_1=2$ and $b_1=a_1$ . From (3.33),

(3.36) $$ \begin{align} P(T)=T^2 + \bigg({-}\dfrac{a_1^2}{2} - 1\bigg)T - \dfrac{a_1^2}{2}.\end{align} $$

Case 2.4: $-d_1/c_1={\zeta _8}^3-{\zeta _8}-1$ . Then, $d_1=-2({\zeta _8}^3-{\zeta _8}-1)$ and

$$ \begin{align*}({\zeta_8}^3-{\zeta_8}-1)a_1+b_1=\pm 2({\zeta_8}^3+{\zeta_8}^2+{\zeta_8}).\end{align*} $$

By changing the signs of $a_1$ and $b_1$ , we can assume that

$$ \begin{align*}b_1=2({\zeta_8}^3+{\zeta_8}^2+{\zeta_8})-({\zeta_8}^3-{\zeta_8}-1)a_1.\end{align*} $$

From (3.33),

(3.37) $$ \begin{align} P(T)& =T^2 + \bigg({-}\dfrac{a_1^2}{2} + ({\zeta_8}^3 - {\zeta_8} - 1)\!\bigg)T + \dfrac{1}{2}({\zeta_8}^3 - {\zeta_8} - 1)a_1^2 \nonumber\\ & \quad + (-2{\zeta_8}^3 - 2{\zeta_8}^2 - 2{\zeta_8})a_1 - 2{\zeta_8}^3 + 2{\zeta_8} + 2. \end{align} $$

Case 2.5: $-d_1/c_1=-({\zeta _8}^3-{\zeta _8}-1)$ . Then, $d_1=2({\zeta _8}^3-{\zeta _8}-1)$ and

$$ \begin{align*}-({\zeta_8}^3-{\zeta_8}-1)a_1+b_1=\pm 2({\zeta_8}^3-{\zeta_8}-1).\end{align*} $$

We can assume that $b_1=2({\zeta _8}^3-{\zeta _8}-1)+({\zeta _8}^3-{\zeta _8}-1)a_1$ . From (3.33),

(3.38) $$ \begin{align} P(T)& = T^2 + \bigg({-}\dfrac{a_1^2}{2} - {\zeta_8}^3 + {\zeta_8} + 1\!\bigg)T + \dfrac{1}{2}(-{\zeta_8}^3 + {\zeta_8} + 1)a_1^2 \nonumber\\ & \quad + (-2{\zeta_8}^3 + 2{\zeta_8} + 2)a_1- 2{\zeta_8}^3 + 2{\zeta_8} + 2. \end{align} $$

Case 2.6: $-d_1/c_1={\zeta _8}^2$ . Then, $d_1=-2{\zeta _8}^2$ and ${\zeta _8}^2a_1+b_1=\pm 2{\zeta _8}^3$ . We can assume that $b_1=2{\zeta _8}^3-a_1{\zeta _8}^2$ . From (3.33),

(3.39) $$ \begin{align} P(T)=T^2 + \bigg({-}\dfrac{a_1^2}{2} + {\zeta_8}^2\bigg)T + \dfrac{1}{2}{\zeta_8}^2a_1^2 - 2{\zeta_8}^3a_1 - 2. \end{align} $$

Case 2.7: $-d_1/c_1=-{\zeta _8}^2$ . Then, $d_1=2{\zeta _8}^2$ and $-{\zeta _8}^2a_1+b_1=\pm 2{\zeta _8}$ . We can assume that $b_1=2{\zeta _8}+{\zeta _8}^2a_1$ . From (3.33),

(3.40) $$ \begin{align} P(T)=T^2 - \bigg(\dfrac{a_1^2}{2}+ {\zeta_8}^2\bigg)T - \dfrac{1}{2}{\zeta_8}^2a_1^2 - 2{\zeta_8} a_1 - 2. \end{align} $$

Case 2.8: $-d_1/c_1={\zeta _8}^3-{\zeta _8}+1$ . Then, $d_1=-2({\zeta _8}^3-{\zeta _8}+1)$ and

$$ \begin{align*}({\zeta_8}^3-{\zeta_8}+1)a_1+b_1=\pm 2({\zeta_8}^3-{\zeta_8}+1).\end{align*} $$

We can assume that $b_1=2({\zeta _8}^3-{\zeta _8}+1)-({\zeta _8}^3-{\zeta _8}+1)a_1$ . From (3.33),

(3.41) $$ \begin{align} P(T)& = T^2 + \bigg({-}\dfrac{a_1^2}{2} + {\zeta_8}^3 - {\zeta_8} + 1\bigg)T + \dfrac{1}{2}({\zeta_8}^3 - {\zeta_8} + 1)a_1^2 \nonumber\\ & \quad + (-2{\zeta_8}^3 + 2{\zeta_8} - 2)a_1 + 2{\zeta_8}^3 - 2{\zeta_8} + 2. \end{align} $$

Case 2.9: $-d_1/c_1=-({\zeta _8}^3-{\zeta _8}+1)$ . Then, $d_1=2({\zeta _8}^3-{\zeta _8}+1)$ and

$$ \begin{align*}-({\zeta_8}^3-{\zeta_8}+1)a_1+b_1=\pm 2({\zeta_8}^3-{\zeta_8}^2+{\zeta_8}).\end{align*} $$

We can assume that $b_1=2({\zeta _8}^3-{\zeta _8}^2+{\zeta _8})+({\zeta _8}^3-{\zeta _8}+1)a_1$ . From (3.33),

(3.42) $$ \begin{align} P(T)& = T^2 + \bigg({-}\dfrac{a_1^2}{2} - {\zeta_8}^3 + {\zeta_8} - 1\bigg)T + \dfrac{1}{2}(-{\zeta_8}^3 + {\zeta_8} - 1)a_1^2 \nonumber\\ & \quad + (-2{\zeta_8}^3 + 2{\zeta_8}^2 - 2{\zeta_8})a_1 + 2{\zeta_8}^3 - 2{\zeta_8} + 2. \end{align} $$

Step 3: One polynomial from (3.24), (3.25), (3.26), (3.27), (3.28), (3.29), (3.30), (3.31) and (3.32) needs to match with one polynomial from (3.34), (3.35), (3.36), (3.37), (3.38), (3.39), (3.40), (3.41) and (3.42), resulting in 81 systems of equations in $a,a_1$ . For each of these systems, MAGMA [2] is used to find a and $a_1$ . MAGMA codes are available from the author on request. Even though $a,a_1\in K$ , each of these 81 systems of equations has coefficients in ${\mathbb {Q}}({\zeta _8})$ , so if a solution with $a,a_1\in K$ exists, then ${a,a_1\in {\mathbb {Q}}({\zeta _8})}$ . Our computation shows that only 20 of these 81 systems have solutions and only 16 of these 20 systems give an irreducible polynomial $P(T)$ . All of the remaining 61 systems have no solutions $a,a_1\in {\mathbb {Q}}({\zeta _8})$ .

Case 3.1: (3.24) and (3.35). Then,

$$ \begin{align*}-\dfrac{a^2}{2}= -\dfrac{a_1^2}{2} + 1,\quad 1=\dfrac{a_1^2}{2}.\end{align*} $$

Thus, $(a,a_1)=(0,\pm \sqrt {2})$ . Hence $P(T)=T^2+1=(T+{\zeta _8}^2)(T-{\zeta _8}^2)$ , which is reducible in $K[T]$ .

Case 3.2: (3.24) and (3.36). Then,

$$ \begin{align*}-\dfrac{a^2}{2}= -\dfrac{a_1^2}{2} - 1,\quad 1=-\dfrac{a_1^2}{2}.\end{align*} $$

Thus, $(a,a_1)=(0,\pm \sqrt {-2})$ . Hence, $P(T)=T^2+1=(T+{\zeta _8}^2)(T-{\zeta _8}^2)$ , which is reducible in $K[T]$ .

Case 3.3: (3.25) and (3.35). Then,

$$ \begin{align*}-\dfrac{a^2}{2} + 1=-\dfrac{a_1^2}{2} + 1,\quad \dfrac{a^2}{2} + 2a + 2=\dfrac{a_1^2}{2}.\end{align*} $$

Thus, $(a,a_1)=(-1,\pm 1)$ . Hence, $P(T)=T^2 + \tfrac 12T + \tfrac 12$ . So $t^2+\tfrac 12t+\tfrac 12=0$ . Therefore,

(3.43) $$ \begin{align} [x:y:z]=[\pm(2t+1):\pm 2t:1].\end{align} $$

Case 3.4: (3.25) and (3.36). Then,

$$ \begin{align*}-\dfrac{a^2}{2} + 1=-\dfrac{a_1^2}{2} - 1,\quad \dfrac{a^2}{2} + 2a + 2=- \dfrac{a_1^2}{2}.\end{align*} $$

Hence, $(a,a_1)=(-2,0),\,(0,\pm 2{\zeta _8}^2)$ . The first solution gives $P(T)=T^2-T$ , which is reducible in $K[T]$ . The second solution gives $P(T)=T^2+T+2$ . So $t^2+t+2=0$ . Therefore,

(3.44) $$ \begin{align} [x:y:z]=[\pm t: \pm (t+1){\zeta_8}^2 :1].\end{align} $$

Case 3.5: (3.25) and (3.39). Then,

$$ \begin{align*}-\dfrac{a^2}{2} + 1=-\dfrac{a_1^2}{2} + {\zeta_8}^2,\quad \dfrac{a^2}{2} + 2a + 2=\dfrac{1}{2}{\zeta_8}^2a_1^2 - 2{\zeta_8}^3a_1 - 2.\end{align*} $$

Hence, $(a,a_1)=({\zeta _8}^2-1,{\zeta _8}+{\zeta _8}^3),\,(-({\zeta _8}^2+3),3{\zeta _8}-{\zeta _8}^3)$ . The first solution gives

$$ \begin{align*}P(T)=T^2 + ({\zeta_8}^2 + 1)T + {\zeta_8}^2=(T+1)(T+{\zeta_8}^2),\end{align*} $$

which is reducible in $K[T]$ . The second solution gives $P(T)=T^2 -(3{\zeta _8}^2 + 3)T + {\zeta _8}^2$ . So $t^2 -(3{\zeta _8}^2 + 3)t + {\zeta _8}^2=0$ . Therefore,

(3.45) $$ \begin{align} [x:y:z]=\bigg[\pm \dfrac{({\zeta_8}^2 - 1)t + {\zeta_8}^2 + 3}{4} : \pm \dfrac{({\zeta_8}^3 + {\zeta_8})(t-3)}{4}:1\bigg].\end{align} $$

Case 3.6: (3.25) and (3.40). Then,

$$ \begin{align*}-\dfrac{a^2}{2}+1=-\dfrac{a_1^2}{2}-{\zeta_8}^2,\quad \dfrac{a^2}{2} + 2a + 2=- \dfrac{1}{2}{\zeta_8}^2a_1^2 - 2{\zeta_8} a_1 - 2.\end{align*} $$

Hence, $(a,a_1)=(-{\zeta _8}^2-1,{\zeta _8}+{\zeta _8}^3),\,({\zeta _8}^2-3,3{\zeta _8}^3-{\zeta _8})$ . The first solution gives

$$ \begin{align*}P(T)=T^2 + (-{\zeta_8}^2 + 1)T - {\zeta_8}^2=(T+1)(T-{\zeta_8}^2),\end{align*} $$

which is reducible in $K[T]$ . The second solution gives $P(T)=T^2 + (3{\zeta _8}^2 - 3)T - {\zeta _8}^2$ . So $t^2 + (3{\zeta _8}^2 - 3)t - {\zeta _8}^2=0$ . Therefore,

(3.46) $$ \begin{align} [x:y:z]=\bigg[{\pm} \dfrac{({\zeta_8}^2 + 1)t + {\zeta_8}^2 - 3}{4}: \pm \dfrac{({\zeta_8}^3 + {\zeta_8})(t-3)}{4}:1\bigg].\end{align} $$

Case 3.7: (3.26) and (3.35). Then,

$$ \begin{align*}-\dfrac{a^2}{2} - 1=-\dfrac{a_1^2}{2} + 1,\quad - \dfrac{a^2}{2} - 2{\zeta_8}^2a + 2=\dfrac{a_1^2}{2}.\end{align*} $$

Hence, $(a,a_1)=(2{\zeta _8}^2,0),\,(0,\pm 2)$ . The first solution gives $P(T)=T^2+T= T(T+1)$ , which is reducible in $K[T]$ . The second solution gives $P(T)=T^2-T+2$ . So $t^2-t+2=0$ . Therefore,

(3.47) $$ \begin{align} [x:y:z]=[\pm t{\zeta_8}^2: (t-1){\zeta_8}^2:1]. \end{align} $$

Case 3.8: (3.26) and (3.36). Then,

$$ \begin{align*}-\dfrac{a^2}{2} - 1=-\dfrac{a_1^2}{2} - 1,\quad - \dfrac{a^2}{2} - 2{\zeta_8}^2a + 2=- \dfrac{a_1^2}{2}.\end{align*} $$

Hence, $(a,a_1)=({\zeta _8}^2,\pm {\zeta _8}^2)$ . Therefore, $P(T)=T^2-\tfrac 12T+\tfrac 12$ . So $t^2-\tfrac 12t+\tfrac 12=0$ . Therefore,

(3.48) $$ \begin{align} [x:y:z]=[\pm (2t-1){\zeta_8}^2: \pm 2t:1].\end{align} $$

Case 3.9: (3.26) and (3.39). Then,

$$ \begin{align*}-\dfrac{a^2}{2} - 1=-\dfrac{a_1^2}{2} + {\zeta_8}^2,\quad - \dfrac{a^2}{2} + 2{\zeta_8}^2 a + 2= \dfrac{1}{2}{\zeta_8}^2 a_1^2 - 2{\zeta_8}^3 a_1 - 2.\end{align*} $$

Hence, $(a,a_1)=({\zeta _8}^2-1,{\zeta _8}-{\zeta _8}^3),\,(3{\zeta _8}^2+1,3{\zeta _8}+{\zeta _8}^3)$ . The first solution gives

$$ \begin{align*}P(T)=T^2 + ({\zeta_8}^2 - 1)T - {\zeta_8}^2=(T-1)(T+{\zeta_8}^2),\end{align*} $$

which is reducible in $K[T]$ . The second solution gives $P(T)=T^2 + (-3{\zeta _8}^2 + 3)T - {\zeta _8}^2$ . So $t^2 + (-3{\zeta _8}^2 + 3)t - {\zeta _8}^2=0$ . Therefore,

(3.49) $$ \begin{align} [x:y:z]=\bigg[{\pm} \dfrac{({\zeta_8}^2 - 1)t +3{\zeta_8}^2 + 1}{4}: \pm \dfrac{({\zeta_8}^3 + {\zeta_8})(t +3)}{4}:1\bigg].\end{align} $$

Case 3.10: (3.26) and (3.40). Then,

$$ \begin{align*}-\dfrac{a^2}{2} - 1=-\dfrac{a_1^2}{2} - {\zeta_8}^2,\quad - \dfrac{a^2}{2} + 2{\zeta_8}^2 a + 2= - \dfrac{1}{2}{\zeta_8}^2 a_1^2 - 2{\zeta_8} a_1 - 2.\end{align*} $$

Hence, $(a,a_1)=({\zeta _8}^2+ 1,{\zeta _8}-{\zeta _8}^3),\,(3{\zeta _8}^2-1,{\zeta _8}+3{\zeta _8}^3)$ . The first solution gives

$$ \begin{align*}P(T)=T^2 + (-{\zeta_8}^2 - 1)T + {\zeta_8}^2=(T-1)(T-{\zeta_8}^2),\end{align*} $$

which is reducible in $K[T]$ . The second solution gives $P(T)=T^2 + (3{\zeta _8}^2 + 3)T + {\zeta _8}^2$ . So $t^2 + (3{\zeta _8}^2 + 3)t + {\zeta _8}^2=0$ . Therefore,

(3.50) $$ \begin{align} [x:y:z]=\bigg[{\pm} \dfrac{({\zeta_8}^2 + 1)t + (3{\zeta_8}^2 - 1)}{4}:\pm \dfrac{({\zeta_8}^3 + {\zeta_8})(t +3)}{4}:1\bigg].\end{align} $$

Case 3.11: (3.27) and (3.34). Then,

$$ \begin{align*}-\dfrac{a^2}{2} + {\zeta_8}^2=-\dfrac{a_1^2}{2},\quad \dfrac{1}{2}{\zeta_8}^2a^2=-1.\end{align*} $$

Hence, $(a,a_1)=(a,a_1)=(\pm \sqrt {2} \zeta _8, 0))$ . Therefore, $P(T)=T^2-1$ , which is reducible in $K[T]$ .

Case 3.12: (3.27) and (3.35). Then,

$$ \begin{align*}-\dfrac{a^2}{2} + {\zeta_8}^2=-\dfrac{a_1^2}{2} + 1,\quad \dfrac{1}{2}{\zeta_8}^2 a^2=\dfrac{a_1^2}{2}.\end{align*} $$

Hence, $(a,a_1)=(\pm ({\zeta _8}+{\zeta _8}^3),\pm ({\zeta _8}^2-1))$ . Therefore, $P(T)=T^2 + ({\zeta _8}^2 + 1)T - {\zeta _8}^2$ . So $t^2 + ({\zeta _8}^2 + 1)t - {\zeta _8}^2=0$ . Therefore,

(3.51) $$ \begin{align}\hspace{-27pt} [x:y:z]=\bigg[{\pm} \dfrac{({\zeta_8}^3 - {\zeta_8})(t+1)}{2}: \pm \dfrac{({\zeta_8}^3 - {\zeta_8})t -{\zeta_8}^3 - {\zeta_8})}{2}:1\bigg].\end{align} $$

Case 3.13: (3.27) and (3.36). Then,

$$ \begin{align*}-\dfrac{a^2}{2} + {\zeta_8}^2=-\dfrac{a_1^2}{2} - 1,\quad \dfrac{1}{2}{\zeta_8}^2 a^2=-\dfrac{a_1^2}{2}.\end{align*} $$

Hence, $(a,a_1)=(\pm ({\zeta _8}^3-{\zeta _8}),\pm ({\zeta _8}^2-1))$ . Therefore, $P(T)=T^2+({\zeta _8}^2-1)T+{\zeta _8}^2$ . So $t^2+({\zeta _8}^2-1)t+{\zeta _8}^2=0$ . Therefore,

(3.52) $$ \begin{align} [x:y:z]=\bigg[{\pm} \bigg(\dfrac{({\zeta_8}^3+ {\zeta_8})(1-t)}{2}\bigg):\pm \dfrac{({\zeta_8}^3 - {\zeta_8})t -{\zeta_8}^3 - {\zeta_8}}{2}:1\bigg].\end{align} $$

Case 3.14: (3.27) and (3.39). Then,

$$ \begin{align*}-\dfrac{a^2}{2} + {\zeta_8}^2=-\dfrac{a_1^2}{2} + {\zeta_8}^2,\quad \dfrac{1}{2}{\zeta_8}^2 a^2=\dfrac{1}{2}{\zeta_8}^2 a_1^2 - 2{\zeta_8}^3 a_1 - 2.\end{align*} $$

Hence, $(a,b)=(\pm {\zeta _8}, {\zeta _8})$ . Therefore, $P(T)=T^2 + \tfrac 12{\zeta _8}^2 T - \tfrac 12$ . So $t^2+\tfrac 12{\zeta _8}^2t-\tfrac 12=0$ . Therefore,

(3.53) $$ \begin{align}[x:y:z]=\bigg[{\pm} \bigg({\zeta_8} t - \dfrac{{\zeta_8}^3}{2}\bigg): \pm \bigg({\zeta_8}^2 t - \dfrac{1}{2}\bigg) :1\bigg].\end{align} $$

Case 3.15: (3.27) and (3.40). Then,

$$ \begin{align*}-\dfrac{a^2}{2} + {\zeta_8}^2=-\dfrac{a_1^2}{2} - {\zeta_8}^2,\quad \dfrac{1}{2}{\zeta_8}^2 a^2=- \dfrac{1}{2}{\zeta_8}^2 a_1^2 - 2{\zeta_8} a_1 - 2.\end{align*} $$

Hence, $(a,b)=(0, 2{\zeta _8}^3),\,(\pm 2{\zeta _8},0)$ . The first solution gives $P(T)=T^2+{\zeta _8}^2T= T(T+{\zeta _8}^2)$ , which is reducible in $K[T]$ . The second solution gives $P(T)= T^2 - {\zeta _8}^2 T - 2$ . So $t^2-{\zeta _8}^2t-2=0$ . Therefore,

(3.54) $$ \begin{align}[x:y:z]=\bigg[{\pm} \bigg(\dfrac{1}{2}{\zeta_8} t - {\zeta_8}^3\bigg):\pm \dfrac{t}{2}:1\bigg]. \end{align} $$

Case 3.16: (3.28) and (3.34). Then,

$$ \begin{align*}-\dfrac{a^2}{2} - {\zeta_8}^2=-\dfrac{a_1^2}{2},\quad - \dfrac{1}{2}{\zeta_8}^2 a^2= -1.\end{align*} $$

Hence, $(a,a_1)=(\pm \sqrt {2} \zeta _8^3, 0)$ . Therefore, $P(T)=T^2-1$ , which is reducible in $K[T]$ .

Case 3.17: (3.28) and (3.35). Then,

$$ \begin{align*}-\dfrac{a^2}{2} - {\zeta_8}^2=-\dfrac{a_1^2}{2} + 1,\quad - \dfrac{1}{2}{\zeta_8}^2 a^2= \dfrac{a_1^2}{2}.\end{align*} $$

Hence, $(a,a_1)=(\pm ({\zeta _8}+{\zeta _8}^3) ,\pm (1+{\zeta _8}^2))$ . Thus, $P(T)=T^2+(1-{\zeta _8}^2)T+{\zeta _8}^2$ . So $t^2+(1-{\zeta _8}^2)T+{\zeta _8}^2=0$ . Therefore,

(3.55) $$ \begin{align}\hspace{-27pt} [x:y:z]=\bigg[{\pm} \dfrac{({\zeta_8}^3 - {\zeta_8})(t+1)}{2}:\ {\pm} \dfrac{({\zeta_8}^3 - {\zeta_8})t + {\zeta_8}^3 + {\zeta_8}}{2}:1\bigg].\end{align} $$

Case 3.18: (3.28) and (3.36). Then,

$$ \begin{align*}-\dfrac{a^2}{2} - {\zeta_8}^2=-\dfrac{a_1^2}{2} - 1,\quad - \dfrac{1}{2}{\zeta_8}^2 a^2= - \dfrac{a_1^2}{2}.\end{align*} $$

Hence, $(a,a_1)=(\pm ({\zeta _8}^3-{\zeta _8}) ,\pm ({\zeta _8}^2+1))$ . Thus, $P(T)=T^2-({\zeta _8}^2+1)T-{\zeta _8}^2$ . So $t^2-({\zeta _8}^2+1)t-{\zeta _8}^2=0$ . Therefore,

(3.56) $$ \begin{align}\hspace{-27pt} [x:y:z]=\bigg[{\pm} \dfrac{({\zeta_8}+{\zeta_8}^3)(t-1)}{2}:\ {\pm} \dfrac{({\zeta_8}^3 - {\zeta_8})t + {\zeta_8}^3 + {\zeta_8}}{2}:1\bigg].\end{align} $$

Case 3.19: (3.28) and (3.39). Then,

$$ \begin{align*}-\dfrac{a^2}{2} - {\zeta_8}^2=-\dfrac{a_1^2}{2} + {\zeta_8}^2,\quad - \dfrac{1}{2}{\zeta_8}^2 a^2= \dfrac{1}{2}{\zeta_8}^2 a_1^2 - 2{\zeta_8}^3 a_1 - 2.\end{align*} $$

Hence, $(a,a_1)=(0,2{\zeta _8}),\, (\pm 2{\zeta _8}^3,0)$ . The first solution gives $P(T)=T^2-{\zeta _8}^2T$ , which is reducible in $K[T]$ . The second solution gives $P(T)=T^2+{\zeta _8}^2T-2$ . So $t^2+{\zeta _8}^2t-2=0$ . Therefore,

(3.57) $$ \begin{align} [x:y:z]=\bigg[{\pm} \bigg(\dfrac{1}{2}{\zeta_8}^3 t - {\zeta_8}\bigg):\pm \dfrac{t}{2} :1\bigg].\end{align} $$

Case 3.20: (3.28) and (3.40). Then,

$$ \begin{align*}-\dfrac{a^2}{2} - {\zeta_8}^2=-\dfrac{a_1^2}{2} - {\zeta_8}^2,\quad - \dfrac{1}{2}{\zeta_8}^2 a^2= - \dfrac{1}{2}{\zeta_8}^2 a_1^2 - 2{\zeta_8} a_1 - 2.\end{align*} $$

Hence, $(a,a_1)=(\pm {\zeta _8}^3,{\zeta _8}^3)$ . Thus, $P(T)=T^2-\tfrac 12{\zeta _8}^2 T-\tfrac 12$ . So $t^2-\tfrac 12{\zeta _8}^2t-\tfrac 12=0$ . Therefore,

(3.58) $$ \begin{align} [x:y:z]=\bigg[{\pm} \bigg({\zeta_8}^3 t - \dfrac{{\zeta_8}}{2}\bigg):\pm \bigg({\zeta_8}^2 t + \dfrac{1}{2}\bigg):1\bigg].\end{align} $$

Note that the 32 points in (3.51), (3.52), (3.55) and (3.56) are the 32 points in Theorem 1.1(C) and the 96 points in (3.43), (3.44), (3.45), (3.46), (3.47), (3.48), (3.49), (3.50), (3.53), (3.54), (3.57) and (3.58) are the 96 points in Theorem 1.1(D).

The proof of Theorem 1.2 is complete.