1 Introduction
For a variety V defined over
$\mathbb {Q}$
, the Hasse principle says that if
$V(\mathbb {R})\neq \emptyset $
and
$V(\mathbb {Q}_p)\neq \emptyset $
for all prime numbers p, then
$V(\mathbb {Q})\neq \emptyset $
. This principle is true for quadratic forms (see Serre [Reference Serre16, Section 3.2, Theorem 8]) but it is not true in general. The classical counterexamples are
$2y^2=1-17x^4$
(Lind [Reference Lemmermeyer10], Reichardt [Reference Reichardt14]),
$3x^3+4y^3+5y^3=0$
(Selmer [Reference Selmer15]) and
$5x^3+9y^3+10z^3+12w^3=0$
(Cassels and Guy [Reference Cassels and Guy5]). For more examples, see Colliot-Thélène et al. [Reference Colliot-Thélène, Kanevsky and Sansuc6], Skorobogatov [Reference Skorobogatov18], Poonen [Reference Poonen12], Quan [Reference Quan13] and Hirakawa [Reference Hirakawa8]. This paper focuses on counterexamples to the Hasse principle in the class of diagonal quartic surfaces
$$ \begin{align} \alpha x^4+\beta y^4+\gamma z^4+\delta w^4=0, \end{align} $$
where
$\alpha ,\,\beta ,\,\gamma ,\,\delta $
are nonzero integers such that
$\alpha \beta \gamma \delta $
is a square. These surfaces were studied extensively by Swinnerton-Dyer [Reference Swinnerton-Dyer19] and Bright [Reference Bright2–Reference Bright4]. However, only a few examples of surfaces (1.1) are known to violate the Hasse principle. The surfaces
$$ \begin{align} 4x^4+9y^4=8z^4+8w^4, \end{align} $$
$$ \begin{align} 2x^4+9y^4=6z^4+12w^4 \end{align} $$
and the family
$$ \begin{align*} x^4+4y^4=d(z^4+w^4), \end{align*} $$
where
$d\in \mathbb {Z}^{+}$
,
$d\equiv 2 \pmod {16}$
, no prime
$p\equiv 3 \pmod 4$
divides d, no prime
$p\equiv 5 \pmod 8$
divides d to an odd power and
$d=r^2+s^2$
with
$r\equiv \pm 3 \pmod 8$
, appeared in Swinnerton-Dyer [Reference Swinnerton-Dyer19]. Bremner and Tho [Reference Bremner and Tho1] found the family
$$ \begin{align} x^4+7P^2y^4=14P^2Q^2z^4+18Q^2w^4, \end{align} $$
where every prime divisor of
$PQ$
is congruent to 1 mod 24, if p is a prime divisor of P, then
$2Q^2$
is a fourth power mod p and, if q is a prime divisor of Q, then
$-7P^2$
is a fourth power mod q. Specialising to
$P=Q=1$
in (1.3) gives the surface
$$ \begin{align} x^4+7y^4=14z^4+18w^4, \end{align} $$
which has a solution
$(2\theta ^2+2\theta ,\, 2\theta ,\, \theta ^2+1,\,\theta ^2-1)$
, where
$\theta ^3+\theta ^2-1=0$
. Currently, the surface (1.4) is the only known example of type (1.1) which violates the Hasse principle but has nontrivial points in a cubic number field. It is an open question whether the surfaces (1.2a) or (1.2b) have points in cubic number fields. It is worth mentioning here the work of Manin [Reference Lind11] and Colliot-Thélène et al. [Reference Colliot-Thélène, Kanevsky and Sansuc6], which is entirely devoted to the study of diagonal cubic surfaces
$$ \begin{align*} \alpha x^3+\beta y^3+\gamma z^3+\delta w^3=0, \end{align*} $$
where
$\alpha ,\beta ,\gamma ,\delta $
are nonzero integers.
The principal results of this paper are Theorems 1.1 and 1.2. For an odd prime number p and an integer a, with
$p\nmid a$
, the symbol
$({a}/{p})_4$
is
$+1$
if a is a fourth power mod p and is
$-1$
otherwise. The symbol
$({a}/{p})$
is defined similarly by replacing fourth powers mod p by squares mod p.
Theorem 1.1. Let
$a,\,b,\,c,\,d$
be square-free integers such that
$-abcd$
is a square,
$a\equiv -b\equiv c\equiv d\equiv \pm 1 \pmod 8$
,
$a\equiv c\equiv d \pmod 3$
,
$({3}/{p} )=({2}/{p} )_4=1$
for any prime divisor p of
$abcd$
and, for any permutation
$(a_1,b_1,c_1,d_1)$
of
$(a,b,c,d)$
and any prime divisor p of
$a_1b_1$
not dividing
$c_1d_1$
, we have
$({c_1d_1}/{p})=1$
. Then the surface
$$ \begin{align} 128a^2x^4+18b^2y^4=c^2z^4+d^2w^4 \end{align} $$
-
(i) is solvable in
$\mathbb {Q}_p$
for all prime numbers p and -
(ii) has no rational points.
Theorem 1.2. Let
$a,\,b,\,c,\,d$
be square-free integers such that
$-abcd$
is a square,
$\gcd (a,b,c,d)=1$
,
$a\equiv -b\equiv c\equiv d\equiv \pm 1 \pmod 8$
,
$a\equiv -b\equiv c\equiv d \pmod 3$
,
$({-1}/{p})=({2}/{p})_4=({3}/{p} )_4=1$
for any prime divisor p of
$abcd$
and, for any permutation
$(a_1,b_1,c_1,d_1)$
of
$(a,b,c,d)$
and any prime divisor p of
$a_1b_1$
not dividing
$c_1d_1$
, we have
$({c_1d_1}/{p})=1$
. Then the surface
$$ \begin{align} 27a^2x^4+24b^2y^4=c^2z^4+2d^2w^4 \end{align} $$
-
(i) is solvable in
$\mathbb {Q}_p$
for all prime numbers p and -
(ii) has no rational points.
To prove Theorems 1.1 and 1.2, we explicitly calculate the Brauer–Manin obstruction following the framework described by Swinnerton-Dyer in [Reference Swinnerton-Dyer19].
The conditions on
$a,b,c,d$
and p in Theorems 1.1 and 1.2 are imposed to guarantee the solubility of (1.5) and (1.6) in
$\mathbb {Q}_p$
for all prime numbers p. There are infinitely many integers
$a,\,b,\,c,\,d$
satisfying these conditions. For example, we mimic the example in [Reference Colliot-Thélène, Kanevsky and Sansuc6, Proposition 5] by letting
$a=1$
,
$b=-1$
and
$c=d=q$
, where q is a prime of the form
$q=r^2+576s^2$
, where
$r,\,s\in \mathbb {Z}^{+}$
. The fact that there are infinitely many prime numbers q of this form follows from Cox [Reference Cox7, Theorem 9.12]. The condition
$({2}/{q} )_4=1$
follows from Silverman [Reference Silverman17, Ch. IX, Proposition 6.6]. From Lemmermeyer [Reference Manin9, page 159], we have
$({-3}/{q} )_4=1$
. Since
$q\equiv 1 \pmod {16}$
, we have
$({-1}/{q} )=({-1}/{q} )_4=1$
. Hence,
$({3}/{q} )_4=1.$
Remark 1.3. When we specialise
$b=-1$
and
$a=c=d=1$
in (1.5) and map
$(x,y,z,w)\mapsto (x/2,y/2,z,w)$
, we have the surface (1.2a). When we specialise
$b=-1$
and
$a=c=d=1$
in (1.6) and map
$(x,y,z,w)\mapsto (x/4,y/2,z,w)$
, we have the surface (1.2b).
For the rest of this paper, for a prime number p,
$(\cdot ,\cdot )_p$
denotes the Hilbert symbol and
$\nu _p(s)$
denotes the highest power of a prime number p dividing s. For a subset S of a field, we set
$S^2=\{x^2\colon x\in S\}$
,
$-S=\{-x\colon x\in S\}$
and
$-3S=\{-3x\colon x\in S\}$
. We need some properties of the Hilbert symbol (see Serre [Reference Serre16, Ch. III]).
-
• Let
$a,\,b,\,c\in \mathbb {Q}_p^{*}$
. Then
$(a,bc)_p=(a,b)_p(a,c)_p.$
-
• Let
$a,\,b\in \mathbb {Q}^{*}$
. Then
$(a,b)_{\infty }\prod _{p\text { prime}}(a,b)_p=1.$
-
• Let q be an odd prime. Let
$a,\,b$
be units in
$\mathbb {Z}_q$
. Then
$(a,b)_q=1.$
-
• Let
$a,\,b\in \mathbb {R}$
. If
$a>0$
, then
$(a,b)_{\infty }=1.$
-
• Let
$a,b\in \mathbb {Q}^{*}$
. Write
$a=p^{\epsilon }a_1$
and
$b=p^{\delta }b_1$
, where
$p\nmid a_1b_1$
. Then and
$$ \begin{align*} (a,b)_p=(-1)^{(a_1-1)(b_1-1)/4+\epsilon(b_1^2-1)/8+\delta(a_1^2-1)/8}\quad \text{if } p=2 \end{align*} $$
$$ \begin{align*} (a,b)_p=(-1)^{\epsilon\delta(p-1)/2}\bigg(\dfrac{b_1}{p} \bigg)^{\epsilon}\bigg(\dfrac{a_1}{p}\bigg)^{\delta}\quad \text{if } p> 2.\end{align*} $$
2 Proof of Theorem 1.1
Lemma 5.2 in Bright [Reference Bright2] implies that Equation (1.5) has solutions in
$\mathbb {Q}_p$
for all primes
$p\not \in \{2,3,5\}$
and
$p\nmid abcd$
. We consider the cases
$p\in \{2,3,5\}$
or
$p \mid abcd$
.
Case 1:
$p=2$
. Since
$$ \begin{align*} a\equiv -b\equiv c\equiv d\equiv \pm 1 \pmod{8}, \end{align*} $$
we have
$|a|,|b|,|c|,|d|\in \mathbb {Z}_2^2$
. Then (1.5) has a solution in
$\mathbb {Q}_2$
, namely
$$ \begin{align*} \bigg(0,\dfrac{1}{\sqrt{|b|}},\dfrac{\sqrt[4]{17}}{\sqrt{|c|}},\dfrac{1}{\sqrt{|d|}}\bigg). \end{align*} $$
Case 2:
$p>2$
. By Hensel’s lemma [Reference Serre16, Section 2.2, Corollary 1], it is enough to show that (1.5) has a nontrivial solution mod p.
-
•
$p=3$
and
$p\nmid abcd$
. Equation (1.5) has a solution
$(1,0,1,1)$
mod
$3$
. -
•
$p=5$
and
$p\nmid abcd$
. Equation (1.5) has a solution:-
–
$(1,1,0,0)$
mod
$5$
if
$5 \mid a^2+b^2$
; -
–
$(0,0,1,1)$
mod
$5$
if
$5 \mid c^2+d^2$
; -
–
$(1,1,\,0,\,1)$
mod
$5$
if
$a^2\equiv b^2\equiv c^2\equiv d^2$
(mod
$5$
); -
–
$(1,0,1,1)$
mod
$5$
if
$a^2\equiv b^2\equiv 1$
(mod
$5$
) and
$c^2\equiv d^2\equiv 4$
(mod
$5$
); -
–
$(1,0,1,1)$
mod
$5$
if
$a^2\equiv b^2\equiv 4$
(mod
$5$
) and
$c^2\equiv d^2\equiv 1$
(mod
$5$
).
-
-
•
$p \mid abcd$
. We only consider the case
$p \mid ac$
and
$p\nmid bd$
. The remaining cases can be proved similarly. Since
$({bd}/{p} )=1$
by the given hypothesis, so there exists
$$ \begin{align*} \bigg(\dfrac{bd}{p} \bigg)=1=\bigg(\dfrac{3}{p}\bigg)=\bigg(\dfrac{2}{p}\bigg)_4=1, \end{align*} $$
$z_0\in \mathbb {Z}$
such that
$18b^2y_0^4\equiv d^2$
(mod p). Therefore, (1.5) has a solution
$(0,y_0,0,1)$
(mod p).
We now show that (1.5) has no rational points. On the contrary, assume that
$(x,y,z,w)$
is a rational point on (1.5). We can further assume that
$x,y,z,w\in \mathbb {Z}$
and
$\gcd (x,y,z,w)=1$
. If
$3 \mid x$
, then considering (1.5) mod
$3$
gives
$3 \mid z$
and
$3 \mid w$
. Hence,
$3 \mid y$
, which is impossible since
$\gcd (x,y,z,w)=1$
. Therefore,
$3\nmid x$
. Similarly,
$3\nmid z$
and
$3\nmid w$
. If
$2 \mid y$
, considering (1.5) mod
$4$
gives
$2 \mid z$
and
$2 \mid w$
. By letting
$z=2z_1$
and
$w=2w_1$
, where
$z_1,\,w_1\in \mathbb {Z}$
, (1.5) reduces to
$8a^2x^4=c^2z_1^4+d^2w_1^4$
, so that
$2 \mid z_1$
and
$2 \mid w_1$
. Hence,
$2 \mid x$
. Therefore,
$2 \mid \gcd (x,y,z,w)$
, which is impossible. Thus,
$2\nmid y$
. Similarly,
$2\nmid z$
and
$2\nmid w$
. So,
$3\nmid xzw$
and
$2\nmid yzw$
.
From (1.5),
$$ \begin{align*} (cz^2-dw^2-6by^2)(cz^2-dw^2+6by^2)=(16ax^2-cz^2-dw^2)(16ax^2+cz^2+dw^2). \end{align*} $$
Therefore, there exist coprime integers
$u,\,v$
such that
$$ \begin{align} u(cz^2-dw^2+6by^2)&=v (16ax^2-cz^2-dw^2), \end{align} $$
$$ \begin{align} v(cz^2-dw^2-6by^2)&=u(16ax^2+cz^2+dw^2). \end{align} $$
Eliminating
$x^2,\,y^2,\,z^2$
and
$w^2$
respectively gives
$$ \begin{align} 6Aby^2+Bcz^2-Cdw^2=0, \end{align} $$
$$ \begin{align} 16Aax^2+Ccz^2+Bdw^2=0, \end{align} $$
$$ \begin{align} 8Bax^2-3Cby^2+Adw^2=0, \end{align} $$
$$ \begin{align} 8Cax^2+3Bby^2+Acz^2=0, \end{align} $$
where
$A=u^2+v^2$
,
$B=u^2+2uv-v^2$
and
$C=u^2-2uv-v^2$
.
Lemma 2.1. For all odd primes p, we have
$(B,6)_p=1.$
Proof. We consider three cases.
Case 1:
$p=3$
. Considering (2.1a) mod 3 gives
$3 \mid v$
. Hence,
$3\nmid u$
. Therefore,
$B\equiv u^2\equiv 1$
(mod
$3$
), so that
$B\in \mathbb {Z}_3^2$
. Hence,
$(B,6)_3=1.$
Case 2:
$p \mid abcd$
. Since
$$ \begin{align*}\bigg(\dfrac{2}{p}\bigg)_4=\bigg(\dfrac{3}{p}\bigg)=1, \quad\mbox{we have}\quad \bigg(\dfrac{6}{p} \bigg)=1.\end{align*} $$
Hence,
$6\in \mathbb {Z}_p^2$
and
$(B,6)_p=1$
.
Case 3:
$p>3$
and
$p\nmid abcd$
.
-
•
$p \mid B$
. Since any common divisor of B and
$AC$
divides the discriminant of
$ABC$
, which is
$-2^{28}$
, we have
$p\nmid AC$
. From (2.2a), (2.3)Therefore,
$$ \begin{align} 6ACbdy^2+BCcdz^2=(Cdw)^2. \end{align} $$
$(6ACbd,BCcd)_p=1$
. Since
$6ACbd$
and
$Ccd$
are units in
$\mathbb {Z}_p$
, we have
$(6ACbd,Ccd)_p=1$
. Thus, from (2.3), (2.4)From (2.2b),
$$ \begin{align} (6ACbd,B)_p=1. \end{align} $$
(2.5)Hence,
$$ \begin{align} -ACac(4x)^2-BCcdw^2=(Ccz)^2.\end{align} $$
$(-ACac,-BCcd)_p=1$
. Since
$-ACac$
and
$-Ccd$
are units in
$\mathbb {Z}_p$
, we have
$(-ACac,-Ccd)_p=1$
. Thus, from (2.5), (2.6)From (2.4) and (2.6),
$$ \begin{align} (-ACac,B)_p=1. \end{align} $$
(2.7)Since
$$ \begin{align} (-6abcd,B)_p=1. \end{align} $$
$-abcd$
is a square, we have
$(-abcd,B)_p=1$
. Therefore, from (2.7),
$(B,6)_p=1$
.
-
•
$p\nmid B$
. In this case, B and
$6$
are units in
$\mathbb {Z}_p$
. Hence,
$(B,6)_p=1.$
Lemma 2.2. We have
$(B,6)_2=-1$
and
$(B,6)_{\infty }=1$
.
Proof. Considering (2.1a) mod
$8$
gives
$-2u \equiv -2v \pmod {8}$
and so
$4 \mid (u-v)$
. Let
$u-v=4k$
, where
$k\in \mathbb {Z}$
. Then
$$ \begin{align*} B=2u^2-(u-v)^2=2(u^2-8k^2). \end{align*} $$
Since
$u^2-8k^2\equiv 1$
(mod
$8$
), we have
$B=2\ell ^2$
, where
$\ell \in \mathbb {Z}_2$
. Hence,
$$ \begin{align*} (B,6)_2=(2,6)_2 =(2,2)_2(2,3)_2 =(-1)^{{(3^2-1)}/{8}} =-1. \end{align*} $$
Since
$6>0$
, it follows that
$(B,6)_{\infty }=1$
.
$$ \begin{align*} (B,6)_{\infty} \times\prod\limits_{p\text{ prime}} (B,6)_p =-1, \end{align*} $$
which contradicts the product formula for the Hilbert symbol. So, the surface (1.5) has no rational points.
3 Proof of Theorem 1.2
The solution of (1.6) in
$\mathbb {Q}_p$
for each prime p is proved in the same way as in Theorem 1.1. We focus on the second part of Theorem 1.2. Assume that
$(x,y,z,w)$
is a rational point on (1.6). Then we can assume that
$x,y,z,w\in \mathbb {Z}$
with
$\gcd (x,y,z,w)=1$
. From (1.6), we also have
$3\nmid zw$
,
$2\nmid zwx$
and
$$ \begin{align*} (9ax^2-cz^2-2dw^2)(9ax^2+cz^2+2dw^2)=2(dw^2-cz^2-6by^2)(dw^2-cz^2+6by^2). \end{align*} $$
Therefore, there exist coprime integers
$u,v$
such that
$$ \begin{align} u(9ax^2-cz^2-2dw^2)&=v(dw^2-cz^2+6by^2), \end{align} $$
$$ \begin{align} v(9ax^2+cz^2+2dw^2)&=2u(dw^2-cz^2-6by^2). \end{align} $$
Eliminating
$x^2,y^2,z^2$
and
$w^2$
respectively gives
$$ \begin{align} 6Aby^2 + Bcz^2 -Cdw^2&=0, \end{align} $$
$$ \begin{align} 9Aax^2 -Ccz^2 -2Bdw^2&=0, \end{align} $$
$$ \begin{align} 3Bax^2 + 2Cby^2 -Adw^2&=0, \end{align} $$
$$ \begin{align} -3Cax^2 + 4Bby^2 + Acz^2&=0, \end{align} $$
where
$A=2u^2+v^2$
,
$B=2u^2 + 2uv - v^2$
and
$C=2u^2 -4uv -v^2$
.
Lemma 3.1. For all primes
$p>3$
:
-
(i)
$(A,6)_p=1;$
-
(ii)
$(C,3)_p=1$
.
Proof. Let
$p>3$
be a prime.
(i) Any common prime divisor of A and
$BC$
divides the discriminant of
$ABC$
, which is
$-2^{28}\cdot 3^{10}$
. Therefore, p is not a common divisor of A and
$BC$
.
Case 1:
$p \mid A$
and
$p\nmid abcd$
. From (3.2b),
$ACac(3x)^2-2BCcdw^2=(Ccz)^2$
. Hence,
$$ \begin{align} (ACac,-2BCcd)_p=1. \end{align} $$
Since
$Cac$
and
$-2BCcd$
are units in
$\mathbb {Z}_p$
, we have
$(Cac,-2BCcd)_p=1$
. Thus, from (3.3),
$$ \begin{align} (A,-2BCcd)_p=1. \end{align} $$
From (3.2c),
$3ABadw^2-6BCaby^2=(3Bax)^2$
. Hence,
$$ \begin{align} (3ABad,-6BCab)_p=1. \end{align} $$
Since
$3Bad$
and
$-6BCab$
are units in
$\mathbb {Z}_p$
, we have
$(3Bad,-6BCab)_p=1$
. Thus, from (3.5),
$$ \begin{align} (A,-6BCab)_p=1. \end{align} $$
$$ \begin{align} (A,3abcd)_p=1.\end{align} $$
Since
$-abcd$
is a square, from (3.7),
$$ \begin{align*}(A,-3)_p=1.\end{align*} $$
Since
$p \mid A=2u^2+v^2$
and
$\gcd (u,v)=1$
, we have
$({-2}/{p} )=1$
. Thus,
$-2\in \mathbb {Z}_p^2$
. Therefore,
$(A,-2)_p=1$
. Hence,
$$ \begin{align*}(A,6)_p=(A,-3)_p(A,-2)_p=1.\end{align*} $$
Case 2:
$p \mid A$
and
$p \mid abcd$
. Then
$$ \begin{align*}\bigg(\dfrac{-1}{p} \bigg)=\bigg(\dfrac{3}{p} \bigg)_4=1 \quad\mbox{and so}\quad \bigg(\dfrac{-3}{p} \bigg)=1.\end{align*} $$
Therefore,
$-3\in \mathbb {Z}_p^2$
and so
$(A,-3)_p=1$
. Since
$p \mid A=2u^2+v^2$
, we have
$( {-2}/{p})=1$
. Hence,
$(A,-2)_p=1$
and
$$ \begin{align*}(A,6)_p=(A,-3)_p(A,-2)_p=1.\end{align*} $$
Case 3:
$p\nmid A$
. Then A and
$6$
are units in
$\mathbb {Z}_p$
. Therefore,
$$ \begin{align*}(A,6)_p=1.\end{align*} $$
(ii) Any common prime divisor of C and
$AB$
divides the discriminant of
$ABC$
, which is
$-2^{28}\cdot 3^{10}$
. Therefore, p is not a common divisor of C and
$AB$
.
Case 1:
$p \mid C$
and
$p\nmid abcd$
. From (3.2a),
$BCcdw^2-6ABbcy^2=(Bcz)^2$
. Hence,
$$ \begin{align} (BCcd,-6ABbc)_p=1. \end{align} $$
Since
$Bcd$
and
$-6ABbc$
are units in
$\mathbb {Z}_p$
, we have
$(Bcd,-6ABbc)_p=1$
. From (3.8),
$$ \begin{align}(C,-6ABbc)_p=1. \end{align} $$
From (3.2b),
$ACacz^2+2ABadw^2=(3Aax)^2$
. Hence,
$$ \begin{align} (ACac,2ABad)_p=1. \end{align} $$
Since
$Aac$
and
$2ABad$
are units in
$\mathbb {Z}_p$
, we have
$(Aac,2ABad)_p=1$
. From (3.10),
$$ \begin{align} (C,2ABad)_p=1. \end{align} $$
$$ \begin{align} (C,-3abcd)_p=1. \end{align} $$
Since
$-abcd$
is a square, from (3.12),
$$ \begin{align*}(C,3)_p=1.\end{align*} $$
Case 2:
$p \mid C$
and
$p \mid abcd$
. Then
$({3}/{p})_4=1$
, so
$({3}/{p})_2=1$
. Therefore,
$3\in \mathbb {Z}_p^2$
and so
$(C,3)_p=1$
.
Case 3:
$p\nmid C$
. Since
$p>3$
, both C and
$3$
are units in
$\mathbb {Z}_p$
. Hence,
$(C,3)_p=1$
.
Lemma 3.2. We have
$(A,6)_2=(C,3)_2=1$
.
Proof. Since
$a\equiv c\equiv d$
(mod
$4$
), taking (3.1a) mod
$4$
gives
$$ \begin{align*} 2u\equiv 2v \pmod{4}. \end{align*} $$
Hence,
$2 \mid (u-v)$
. Therefore, u and v are odd. Thus,
$A=2u^2+v^2\equiv 3 \pmod 8$
and
$C=2(u-v)^2-3v^2\equiv 1 \pmod 4$
. Let
$A=8h+3$
and
$C=4h_1+1$
, where
$h,h_1 \in \mathbb {Z}$
. Then
$$ \begin{align*} (A,6)_2=(8h+3,6)_2 =(-1)^{{(8h+3-1)(3-1)}/{4}+{(8h+3)^2-1}/{8}} =1 \end{align*} $$
and
$$ \begin{align*} (C,3)_2=(4h_1+1,3)_2 =(-1)^{{(4h_1+1-1)(3-1)}/{4}} =1. \end{align*} $$
This completes the proof.
Lemma 3.3. We have
$A\in \mathbb {Q}_3^2$
or
$A\in -3\mathbb {Q}_3^2$
. Furthermore, if
$A\in \mathbb {Q}_3^2$
, then
$C\in -\mathbb {Q}_3^2$
.
Proof. We consider two cases.
Case 1:
$3 \mid uv$
.
If
$3 \mid u$
, then
$3\nmid v$
. Since
$A\equiv v^2$
(mod
$3$
) and
$C\equiv -v^2$
(mod
$3$
), it follows that
$A\in \mathbb {Q}_3^2$
and
$C\in -\mathbb {Q}_3^2$
.
Otherwise,
$3 \mid v$
and
$3\nmid u$
. Since
$c\equiv d$
(mod
$3$
), we have
$9 \mid z^2+2w^2$
from (3.1a) and
$9 \mid w^2-z^2-6y^2$
from (3.1b). Therefore,
$9 \mid 3w^2-6y^2$
, so that
$3 \mid w^2-2y^2$
, which is impossible.
Case 2:
$3\nmid uv$
. Then
$3\nmid u$
and
$3\nmid v$
. Hence,
$3 \mid A$
.
Case 2.1:
$3\nmid u-v$
. Then
$$ \begin{align*} B=3u^2-(u-v)^2\equiv -1 \pmod{3}, \end{align*} $$
$$ \begin{align*} C=2(u-v)^2-3v^2\equiv -1\pmod{3}. \end{align*} $$
Therefore,
$B=-\beta ^2$
and
$C=-\gamma ^2$
, where
$\beta ,\,\gamma \in \mathbb {Z}_3$
and
$3\nmid \beta \gamma $
. Then (3.2c) and (3.2d) become
$$ \begin{align} 3\beta^2 x^2+2\gamma^2 y^2+Aw^2&=0 , \end{align} $$
$$ \begin{align} 3\gamma^2 x^2-4\beta^2 y^2+2Az^2&=0. \end{align} $$
Since
$3\nmid \beta \gamma $
and
$3 \mid A$
, from (3.13a) and (3.13b),
$3 \mid y$
. Let
$y=3y_1$
, where
$y_1\in \mathbb {Z}$
. Then (3.13a) gives
$Aw^2=-3(\beta ^2x^2+6\gamma ^2y_1^2).$
Since
$3\nmid \beta x$
, we have
$A\in -3\mathbb {Q}_3^2$
.
Case 2.2:
$3 \mid u-v$
. Let
$u-v=3t$
, where
$t\in \mathbb {Z}$
. Then
$$ \begin{align*} B=3u^2-(u-v)^2=3(u^2-3t^2), \end{align*} $$
$$ \begin{align*}C=2(u-v)^2-3v^2=3(6t^2-v^2).\end{align*} $$
Therefore,
$B=3\beta ^2$
and
$C=-3\gamma ^2$
, where
$\beta ,\gamma \in \mathbb {Z}_3$
and
$3\nmid \beta \gamma $
. Equation (3.2a) becomes
$$ \begin{align} 6A_1by^2+\beta^2 cz^2+\gamma^2 dw^2=0, \end{align} $$
where
$A_1=A/3$
. Since
$c\equiv d$
(mod
$3$
) and
$3\nmid \beta \gamma cdzw$
, (3.14) is impossible mod 3.
Lemma 3.4. We have:
-
(i)
$(C,3)_3=-1$
if
$A\in \mathbb {Q}_3^2$
; -
(ii)
$(A,6)_3=-1$
if
$A\in -3\mathbb {Q}_3^2$
; -
(iii)
$(A,6)_{\infty }=(C,3)_{\infty }=1$
.
Proof. (i) If
$A\in \mathbb {Q}_3^2$
, then
$C\in -\mathbb {Q}_3^2$
. Therefore,
$$ \begin{align*}(C,3)_3=(-1,3)_3=-1.\end{align*} $$
(ii) If
$A\in -3\mathbb {Q}_3^2$
, then
$$ \begin{align} (A,6)_3=(-3,6)_3 =(-3,2)_3(-3,3)_3 =-1. \end{align} $$
(iii) Since
$6>0$
and
$3>0$
,
$$ \begin{align*} (A,6)_{\infty}=(C,3)_{\infty}=1.\\[-36pt] \end{align*} $$
Proof of Theorem 1.2
By Lemma 3.3, we need to consider two cases.
Case 1:
$A\in \mathbb {Q}_3^2$
. Then
$C\in -\mathbb {Q}_3^2$
. From Lemmas 3.1, 3.2 and 3.4,
$$ \begin{align} (C,3)_{\infty}\times \prod\limits_{p\text{ prime}}(C,3)_p=-1. \end{align} $$
Case 2:
$A\in -3\mathbb {Q}_3^3$
. From Lemmas 3.1, 3.2 and 3.4,
$$ \begin{align} (A,6)_{\infty}\times \prod\limits_{p\text{ prime}}(A,6)_p=-1. \end{align} $$
Both (3.16) and (3.17) contradict the product formula for the Hilbert symbol. So, the surface (1.6) has no rational points.
Acknowledgments
The author is sincerely grateful to the referee for many valuable comments and suggestions, especially in the proof of Lemma 3.3, which make the current version of this paper much shorter than the previous one. Part of this work was finished during the author’s stay at the Vietnam Institute of Advanced Study in Mathematics (VIASM). The author would like to thank the Institute for their support.
