数理解析研究所講究録998 巻 1997 年 93-101

93

BINARY CUBIC FORMS WITH MANY INTEGRAL POINTS(prepared in

ERIC LIVERANCE
with Cameron Stewart)$\mathrm{C}\mathrm{o}\mathrm{u}\mathrm{a}\mathrm{b}_{\mathrm{o}\mathrm{r}\mathrm{a}\mathrm{t}}\mathrm{i}_{0}\mathrm{n}$ABSTRACT. We make more explicit a result of Silverman’s on integral points of binary
cubic forms. Using the explicit 3-descent of Satg\’e, we show how to construct suchbinary cubic forms. As an application, we use Quer’s elliptic curves with j-invariant
and rank 12 over , to construct binary cubic forms which have many integral
points.
$0$$\mathbb{Q}$INTRODUCTION
Let $F$ be an irreducible binary form with integer coefficients and degree, the equation
Then for$\geq 3$.

$m\in \mathbb{Z}$$F(U, V)=m$

with $U,$ $V$ integral is known as Thue’s equation. Many studies have been made of
the problem of estimating the number of solutions, which we denote $N_{F}(m)$ , andwhich Thue showed in 1909 must be finite. In 1935, Mahler [Ma] showed that forsome constant $C>0$ independent of $m$$N_{F}(m)>c(\log|m|)^{1/}4$

for infinitely many $m$ . In 1983, Silverman [Sill] improved Mahler’s exponent to 1/3
in general by the following theorem.

Theorem (Silverman). Let $F(U, V)\in \mathbb{Z}[U, V]$ be a cubic form with non-zero disbe an integer such that the curve $E:F(U, V)=m_{0}W^{3}$ has
criminant. Leta point defined over Q. Using that point as an origin, we may give $E$ the structureof an elliptic curve with rank R. Then there is a constant $C>0$ independent of $m$such that$m_{0}\in \mathbb{Z}$$N_{F}(m)>C(\log|m|)R/(R+2)$

for infinitely many$m$.

To obtain the exponent 1/3 from the theorem, Silverman uses a trick to produce
twists of an arbitrary binary cubic form with rank at least one.
A natural question arising in this context is what type of elliptic curves are in: $y^{2}=x^{3}+D$ .
the form in Silverman’s theorem? Let$E_{D}$

Typeset by$A_{\mathrm{A}4}\mathrm{S}\mathrm{I}\mathrm{E}\mathrm{x}$94

Proposition 1. Let $F(U, V)\in \mathbb{Z}[U, V]$ be a cubic form with non-zero discriminant$D$. If the curve $C$ : $mW^{3}=F(U, V)$ has a mtional point overand letthen is birationally isomorphic over to: $y^{2}=x^{3}+m^{2}D/4$ .
Remark 1. The existence of the automorphism : $(U, V, W)\vdasharrow(U, V, \rho W),$ $\rho=$, is enough to show that the -invariant ofis , or equivalently, that$y^{2}=x^{3}+D$. To see this, just take a point of overmay be put in the form
some extension of with $W=0$ as the origin. Then is an endomorphism oforder 3, i.e., an automorphism. As the ring of endomorphisms must be an orderin an imaginary quadratic field, we see that the only ring of endomorphisms (overnumber fields) with this property is. Furthermore, the curves $y^{2}=x^{3}+D$ are
the only elliptic curves with this endomorphism ring.
Remark 2. Note that if the choice of origin is fixed, the map may not be an isogeny
in that it may not fix the origin. In general, any algebraic map between elliptic
curves is the composition of an isogeny (or endomorphism in this case) followed bytranslation by a point (Silverman [Si12] III.4.4),. Now, $\eta^{3}=1$ impliesthat$m\in \mathbb{Z}$$\mathbb{Q}_{f}$$C$$E_{m^{2}D/4}$$\mathbb{Q}$$\eta$$e^{2\pi i/3}$

$C$

$j$

$C$

$0$

$C$

$\mathbb{Q}$$\eta$$\mathbb{Z}[p]$$\eta$$\eta=\tau_{Q^{\circ}}\zeta$$P=\zeta(\zeta(\zeta P+Q)+Q)+Q)=\zeta^{3}P+\zeta(\zeta 2\zeta++1)Q$.

As this is true for every $P$ , we find that is a cube root of unity. Moreover, if$\zeta=1$ , thenmust be a three-torsion point. For example, consider the curve$\zeta$$Q$$C:W^{3}=2S(U3-3rU2V-4DV^{3})$where $s^{2}=r^{3}+D$ (see (6) below). If we take the rational point $(2r, 1, -2s)$ as theorigin of $C$ , then one can show that on $y^{2}=x^{3}-27D$ (and under the isomorphism
given in (10) below), acts as multiplication by $\rho((x, y)rightarrow(\rho x, y))$ followed bytranslation by the point $S=(-3rp^{2},3s(2\rho+1))$ :$\psi$$\eta$$Crightarrow C\eta$$\psi\downarrow$$\downarrow\psi$$E’arrow\tau_{S}0\rho E’$

Lemmal. If : $Earrow E’$ is any non-constant map between elliptic curves of degree($i.e.$ , amap) defined over , then $E=E’/G$ , where is a Galois-invariant
subgroup of $dd$ -torsion points.
$\psi$$d$$d_{-}\iota \mathit{0}-l$$G$$\mathbb{Q}$

Proof of Lemma. Any algebraic map th between elliptic curves can be written as, where is translation on $E’$ and, we have the relation$\psi=\tau\circ\phi$$\mathbb{Q}$$\tau$$\phi$

is a -isogeny. As$d$$\psi$

is defined over,
, we have$\tau\circ\emptyset=\psi=\psi\sigma=\mathcal{T}\circ\phi\sigma\sigma$

where a is any Galois automorphism.

$\mathrm{A}\mathrm{p}\mathrm{p}\mathrm{l}\mathrm{y}\mathrm{i}\mathrm{n}\mathrm{g}-\mathcal{T}\sigma$$-\mathcal{T}^{\sigma_{\mathrm{O}}}\mathcal{T}\circ\emptyset=\emptyset\sigma$

Now if.

is translation by a point $P$ , and we apply the above map to , we find$P-P^{\sigma}=O$ , or
. This plies
and hence. As
is Galois
aariant, it isover Q. Hence, so is he dual isogeny (useuniqueness of the dual isogeny with respect to. Hence $C$ is isomorphic
to
.
$O$

$\tau$$\tau^{\sigma}=\tau$$\mathrm{i}\mathrm{n}\dot{\mathrm{v}}$$\mathrm{i}\grave{\mathrm{m}}$$\phi=\phi^{\sigma}$$\dot{\mathrm{t}}$$\mathrm{k}\mathrm{e}\mathrm{r}\phi$$\hat{\phi}\circ\emptyset=[d])$$E/\mathrm{k}\mathrm{e}\mathrm{r}\hat{\phi}$$\square$$\mathrm{k}\mathrm{e}\mathrm{r}\phi=\mathrm{k}\mathrm{e}\mathrm{r}\phi^{\sigma}$$\dot{\mathrm{d}}\mathrm{e}\mathrm{f}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{d}$$\hat{\phi}$

95

Lemma 2. Let: $y^{2}=x^{3}+D$ . The only Galois invariant subgroupof three
3-torsion points ofis $\{O, (0, \pm\sqrt{D})\}$ . Hence the only elliptic curve thatis3-isogenous to over is $y^{2}=x^{3}-27D$ .
$E_{D}$$G$$E_{D}$$E_{D}$$\mathbb{Q}$

Proof. The 3-torsion points of$E_{D}$

are$E_{D}[3]=\{O, (\mathrm{o}, \pm\sqrt{D}), (-\rho\sqrt{4D}3, \pm\sqrt{-3D}j)\}$,where $p=e^{2\pi i/3}$ is a cube root of unity. See Velu [Ve] to prove the second part.

Proof of Proposition 1. Again, we note that since$\square$has a rational point, it is anelliptic curve. We shall produce a non-constant rational map of degree 3 from $C$
to $E_{-27m^{2}D/4}$ . Then by the lemmas, we can conclude that $C$ is isomorphic to$C$.
As for the degree 3 map, write explicitly$E_{729m^{2}D/D/}4\simeq E_{m}24$$F(x, y)=ax^{3}+byx^{2}+w^{2_{X}}+dy^{3}$ .

Then let [Mo] $H$ be the quadratic$H(x, y)=|_{\frac{\frac{\partial^{2}F}{\partial^{2}F\partial x^{2}}}{\partial x\partial y}}$variant of $F$$\mathrm{c}.0$$\frac{\partial^{2}F}{\frac{\partial x\partial y\partial^{2}F}{\partial y^{2}}}|=(b_{X+y}c)^{2}-(3aX+by)(cx+3dy)$$=(b^{2}-3ac)X2+(bc-9ad)_{X}y+(c^{2}-3bd)y^{2}$and let $G$ be the cubic covariant of $F$ given by$G(x, y)=|_{\frac{\frac{\partial F}{\partial H\partial x}}{\partial x}}$$\frac{\frac{\partial F}{\partial H\partial y}}{\partial y}|=(-27da^{2}+9Cba-2b^{3})X^{3}+(-27dba+18ca-32cb2)yX^{2}$$+(27dca-18db^{2}+3C^{2}b)y2_{X}+(27d^{2}a-9dcb+2c^{3})y^{3}$.

Then we have the relation$(G/2)^{2}=H \mathrm{s}-\frac{27}{4}DF^{2}$,

where $D$ is the discriminant of $F$ . Thus

(1)

$x=H(U, V)/W^{2}$ ;$y=G(U, V)/2W^{3}$

gives a map from $C$ : $mW^{3}=F(U, V)$ to $E_{-27m^{2}D/}4$ (note that $F(U, V)/W^{3}=$$m)$ .
$\square$

SATG\’E’S EXPLICIT 3-DESCENT
We use descent theory to construct binary cubic forms associated to elliptic
curves of the form $y^{2}=x^{3}+D$ . First we recall some facts about the generaldescent theory of elliptic curves (see [Si12] Chapter X) and from Satg\’e’s 2 papers
on explicit 3-descent for elliptic curves with -invariant (see
).
$j$

$0$

$[\mathrm{S}\mathrm{a}\mathrm{t}\mathrm{l}],[\mathrm{s}\mathrm{a}\mathrm{t}2]$

96

To set notation, let $E:y^{2}=x^{3}+D$ where $D$ is a non-zero 6th power free integer.
Then (see [Ca]) there is a -isogeny $\emptyset:Earrow E’$ , where $E’$ : $y^{2}=x^{3}-27D$ and also
the dual isogeny : $E’arrow E$ defined by$\hat{\phi}$$\hat{\phi}(x,y)=(\frac{x^{3}-108D}{9x^{2}},$$y \frac{x^{3}+216D}{27x^{3}})$.

Letbe the absolute Galois group of Q. Then from the exact sequence of Gmodules$G$$0arrow E’[\hat{\phi}]arrow E’arrow E\hat{\emptyset}arrow 0$,

we take Galois homology and from the long exact sequence, we obtain$0arrow E(\mathbb{Q})/\hat{\phi}(E’(\mathbb{Q}))-\delta H1(G, E’[\hat{\emptyset}])arrow H^{1}(G, E’)[\hat{\phi}]arrow 0$(2)$P\mapsto\{\sigma\mapsto Q^{\sigma}-Q\}$,

where is any point satisfying $\hat{\phi}(Q)=P$ and is the connecting homomorphism.
Now there is a bijective correspondence between cocycles inand curves$C$ which are$E’$
twists of(modulo an equivalence relation) (see [Si12] Theorem 3.6,
p. 291). Satg\’e explicitly calculates a curve $C$ whose class corresponds to the cocycle
as follows.
$Q$$\delta$$H^{1}(G, E’[\hat{\phi}])$$\delta(P),$$P\in E(\mathbb{Q})$

Theorem (Satg\’e). Let
be the Galois closure of the field of definition of .
Then contains
as subfield withcyclic cubic. If
is non-abelian of degree 6. Write $-27D=D_{1}D_{2}^{2}$ withthen$p(x)$
square-free. If
is any cubic polynomial whose splitting field is , then$d_{\dot{i}}sc(p)--D1m2$
some
integer $m$ . Then the curve
for
$L$

$L$

$Q$

$\mathbb{Q}(\sqrt{-27D})$$\mathbb{Q}(\sqrt{-27D})_{f}$$L/\mathbb{Q}(\sqrt{-27D})$$\mathbb{Q}\neq$$L/\mathbb{Q}$$D_{1}$$L$$W^{3}=(2D_{2})^{2}mp(U, V)$

comresponds to the cocycle $\delta(P)=\{\sigma\mapsto Q^{\sigma}-Q\}$ in (2) above, where $p(U, V)$ is thehomogenization of $p(x)$ .

Now given a point $P=(r, s)$ on $E$ , we can explicitly construct the curve $C$ whichcorresponds to it under (2). If $Q=(u, v)$ , then we must solve $\hat{\phi}(Q)=P$ , or$\frac{u^{3}-108D}{9u^{2}}=r$

But the second equation gives
equation is equivalent to$v$$v \frac{u^{3}+216D}{27u^{3}}=s$.

rationally in terms of everything else. The first$u^{3}-9ru-2108D=0$ .

Hence the field$L$in the theorem is the splitting field of the polynomial $p(x)=$Now$x^{3}-3rx^{2}-4D$ , where we have absorbed a factor of 3.

.

$\mathrm{d}\mathrm{i}\mathrm{S}\mathrm{c}(p)=-2^{4}3^{3}D(r^{3}+D)=-2^{4}3^{3}Ds^{2}=-3^{3}D(24_{S^{2}})=D_{1}(D_{2}2^{2}s)^{2}$

97Hence, by Satg\’e’s theor.em, $m=2^{2}D_{2}s$ and our curve becomes$W^{33}=16D2ps(U, V)$ .

By absorbing cubes, we can rewrite this as$C:W^{3}=(2s)p(U, V)=2S(U3-3rU2V-4DV^{\mathrm{s}})$ .

Here the discriminant of the binary cubic form $F(U, V)=(2s)p(U, V)$ is$\mathrm{d}\mathrm{i}_{\mathrm{S}}\mathrm{c}(2Sp(U, V))=-108D(2_{S})^{6}$.

If $H(U, V)$ and $G(U, V)$ are the quadratic and cubic covariants of $F$ respectively
then (as above) we have a map of $C$ to $E$ by$x= \frac{H(U,V)}{2^{2}3^{2_{S^{2}}}W^{2}}=(r^{2}U^{2}-32sDVU+32rsDV^{2})/W2$(3)$y=- \frac{G(U,V)}{2^{4}3^{3}S^{3}W^{3}}=-((\Gamma^{\mathrm{s}}+2D)U3-6rDVU2+12r^{2}DV^{2}U+8D^{2}V^{3})/W^{3}$.

Note that this minus sign is forced on us in order that (5) below have solutions(The reason for the minus sign is from descent theory. See [Si12] p. 292, [Ca] or [Li]and also see (11) below.). Setting $u=U/W$ and $\eta’=V/W$ , we must solve(4)$\frac{H(u,v)}{2^{2}3^{2}s^{2}}=r$,$\frac{G(u,v)}{2^{4}3^{3_{S}3}}=-S$,$F(u, v)=1$

or$4Drv^{2}-4Duv-r^{22}u+r=0$(5)$8D^{2}v^{3}+12Dr^{2}uv^{2}-6Druv2+r^{3}u^{3}+2Du^{3}+s=0$$2s(u^{3}-3ruv-42Dv^{3})=1$Eliminating in the first two equations (that is, taking the resultant with respectto ) we find that must satisfy$16r^{\mathrm{s}}S^{6}u-624rS54u^{4}+8r^{3_{S}53}u+9\Gamma^{7}Su-226rS^{3}5u+r^{6}(r^{3}+2D)=0$ ,from which it is easy to find the root $u=-r/s$ . Substituting into either of theequations above, we find $v=-1/2s$ .
Proposition 2. If $P=(r, s)$ is a point on $E(\mathbb{Q})_{f}E:y^{2}=x^{3}+D$ then
$v$

$v$

$u$

(6)

$C:W^{3}=2s(U^{3}-3rU^{2}V-4DV^{\mathrm{s}})$has the rational point $(U, V, W)=(2r, 1, -2s)$ . Moreover $C$ is isomorphic over
to $E’\wedge\cdot y^{2}=x^{3}-27D$ .

$\mathbb{Q}$

Proof. To prove the second statement, we can use standard facts from descent
under the correspondence between thetheory. First note $C$ corresponds toWeil-Chatelet group $WC(E’)$ and $H^{1}(G, E’)$ (Theorem 10.3.6). By [Si12]
Proposition 10.3.3,.
implies that$\delta(P)$$[\mathrm{S}\mathrm{i}12]$$C(\mathbb{Q})\neq\emptyset$$C\simeq_{\mathbb{Q}}E’$$\square$Using this point and the above calculation with Silverman’s theorem, we obtainthe following result.
Theorem. Suppose that: $y^{2}=x^{3}+D$ is of $mnkR$ and that $(r, s)$ is a point$F(U,V)=2s(U^{32}-3rUV-4DV^{\mathrm{s}})$. Let. Then there is a constant $C>0$
on
independent of $m$ such that
$E_{D}$$E_{D}$$N_{F}(m)>C(\log|m|)R/(R+2)$

for infinitely many .
$m$

98

$\mathrm{Q}\mathrm{U}\mathrm{E}\mathrm{R}’ \mathrm{S}$CURVES

Now using Quer’s curves of high rank, we can produce binary cubic forms withmany integral points. To, note that Quer [Q] finds 3 curves of this form
with rank 12, namely $y^{2}=x^{3}+D_{i}$ where$\mathrm{i}\mathrm{u}_{\mathrm{u}}\mathrm{S}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}$$D_{1}=-6533891544658786928$$D_{2}=-49317122354452517296$$D_{3}=-50586546986138596528$ .

The following points$P_{i}$

are on $y^{2}=x^{3}+D_{i}$ :$P_{1}=(2109824, 1690470036)$$P_{2}=(3676420, 611232948)$$P_{3}=$$(3706924, 592751364)$ .

By the calculations above, these correspond to the homogeneous spaces$C_{1}$$C_{2}$$C_{3}$: $W3=338094\mathrm{o}\mathrm{o}72(U^{3}-6329472U^{2}V+26135566178635147712V^{\mathrm{s}})$: $W\mathrm{s}=1222465896(U3-1102926\mathrm{o}U2V+19726848941781\mathrm{o}\mathrm{o}69184)$: $W^{3}=1185502728(U3-11120772U^{2}V+202346187944554386112)$ .

Hence these forms satisfy the following result: for infinitely many integers $m$ ,the number of solutions to $2s(U^{3}-3rU^{2}V-4DV^{\mathrm{s}})=m$ in integer $V$ is greaterthan$C(\log|m|)^{6/}7$ ,$U,$

for some positive constant $C$ .
THE EXPLICIT ISOMORPHISM

Either of Propositions 1 and 2 tells us that since$c:W^{\mathrm{s}_{=2s}3}(U-3rU2V-4DV3)$

has the rational point $O=(U, V, W)=(2r, 1, -2s)$ and discriminant $-108D(2s)^{6}$ ,it should be isomorphic over to $E’$ : $y^{2}=x^{3}-27D,$ : $Carrow E’\sim$ (in the notationof the proposition, we have $m=1$ and we have absorbed the 6th powers intoand ). This amounts to taking this rational point as the origin, and putting $C$
in Weierstrass form in the standard way using the Riemann Roch Theorem (seeSilverman’s Proposition 3.3.1 [Si12] . Thus we must find functions and whose
only poles are atand of order 2 and 3 respectively. An explicit way to obtainthese functions is as follows ([ST], but see also Chapter 10 of [Mo]). We let be
the line tangent to: $2sV+W=0$ . Thenintersects $C$ at another point$P=(r, -1,2S)$ . We let be the line tangent to: $3r^{2}U-(r^{3}+4D)V-2SW=0$$C$
which meets at a third point$\psi$$\mathbb{Q}$

$x$

$y$

$)$

$x$

$y$

$O$

$\hat{Z}$$O,\hat{Z}$$\hat{X}$$\hat{Z}$$P,\hat{X}$$Q=[-r(27r-6108s23r+80s^{4}), 27r^{6}-16s,4S(4636s27r-23r+8s^{4})]$ .

99

Finally, we take to be any other line through$C$ also at
the points and . Now note that$\hat{Y}$$R$$(\hat{Z})=2(o)+(P)-I$ ;

where
let

$I$

$O$

, say

$\hat{Y}$: $sU+rW=0$ , meeting$S$$(\hat{X})=2(P)+(Q)-I$ ;$(\hat{\mathrm{Y}})=(O)+(R)+(s)-I$,

is the formal sum of the 3 points at infinity (i.e. $W=0$ ) on $C$ . Now, if weand, then$\hat{x}=\frac{X}{Z}$$\hat{y}=\frac{Y}{Z}$$(\hat{x})=(\hat{X})-(\hat{Z})=(P)+(Q)-2(o)$ ;$(\hat{y})=(\hat{Y})-(\hat{Z})=(R)+(s)-(P)-(O)$ .

Then using the notation of Riemann-Roch, we find that$\mathcal{L}(4(O)+(P))$ , but yet $\mathcal{L}(4(O)+(P))$ is a 5-dimensional space. Hence wemustobtain a relation of the form$\{1, \hat{x},\hat{x}^{2},\hat{y},\hat{x}\hat{y},\hat{x}\hat{y}^{2}\}\subset$$\hat{x}\hat{y}^{2}+(a\hat{x}+b\hat{y})g=c\hat{x}^{2}+d\hat{x}+e$.

Remark. Note that if the coefficientoris zero then each function has adifferent order of vanishing at , hence all would be zero. Similarly, if the coefficientof is zero, then the coefficient ofmust be as well as it is thether function
with a pole at $P$ . Thus we find $bc\neq 0$ .
This translates into a set of 9
equations in the unknowns. Solvingli.nearthese equations, we find that$\hat{x}^{2}$$\hat{x}\hat{y}^{2}$$O$$\hat{x}\hat{y}^{2}$$\hat{y}$$\mathrm{o}\mathrm{n},1\mathrm{y}.0$$a,$ $b,$ $c,$ $d,$ $e$$a= \frac{-2r^{3}+4D}{3r^{2}}$$b= \frac{s(-r+83D)}{3r^{2}}$$2s^{3}$$c=\overline{9r^{4}}$$d= \frac{8s^{4}+3r^{32}s-9r6}{9r^{4}}$$e= \frac{s(16S^{4}+12_{\Gamma^{3}}S^{2}-27r)6}{18r^{4}}$Multiplying by$\hat{x}$and setting(7)$y=\hat{x}\hat{y}$and$x=\hat{x}$, we find$y^{2}+(ax+b)y=cx^{3}+dx^{2}+ex$ .

And now we have$(x)=(P)+(Q)-2(o)$ ;$(y)=(Q)+(R)+(s)-3(o)$ .

Explicitly, we have
$x= \frac{3r^{2}U-(r+4\mathrm{s}D)V-2sW}{2sV+W};y=\frac{(_{SU+rW})(3r^{2}U-(r+4D)3V-2sW)}{(2sV+W)^{2}}$100and projectively,$X=(3r^{2}U-(r^{3}+4D)V-2SW)(2sV+W)$(8)$\mathrm{Y}=(sU+rW)(3r^{2}U-(r^{3}+4D)V-2sW)$$Z=(2_{S}V+W)^{2}$

Remark. Note that this map is not regular at $P$ or at . One may define a map
compatible with the map above which is regular at these points. To see this,, where is any other line through . By
above bymerely multiply
examining divisors, one may verify that goes to $[0,0,1]$ and that $P$ is the otherpoint lyin on the line $x=0$ , hence $P$ goes to$O$$\ell/\hat{X}$$X,\mathrm{Y},\mathrm{a}\mathrm{n}\mathrm{d}Z$$Q$$\ell$$Q$$\mathrm{g}$$[0, -b, 1]=[0, s(r^{3}-8D), 3r]2$ .

By construction, we know thatmust go to $[0,1,0]$ . In any case, we obtain an
isomorphism of algebraic varieties.
Finally, to put (7) in the appropriate Weierstrass form, we (I) complete thesquare on the left; (II) scale and so that the lead coefficients on left and rightare both equal to 1; (III) translate to remove the -term (and in this case thisalso removes the -term); (IV) remove sixth powers from the constant term. In
other words, if we make the substituti.ons
$O$

$x$

$y$

$x^{2}$$x$$x$$x’= \frac{(3r)^{2}}{s^{2}}(CX+\frac{d+a^{2}/4}{3})$$y’= \frac{(3r)^{3}}{s^{3}}c(y+\frac{1}{2}(ax+b))$

then$\mathrm{w}.\mathrm{e}$find that$y^{\prime 2}=x-\prime 372D$.

Projectively, we find
$X’=r(2SX+(4s^{2}-3r^{\mathrm{s}})z)$(9)$\mathrm{Y}’=6r^{2}\mathrm{Y}+(-2r^{3}+4D)X+(s(-r^{3}+8D))Z$$Z’=r^{3}Z$

where now
$Z/\mathrm{Y}\prime 2=x\prime 33-27DZ’$.

Again, we note that goes to $[0,1,0],$ $P$ goes to $[r(4s^{2}-3r3), s(9r-38s^{2}).’ r]3$ andgoes to $[r(4s^{2}-3r\mathrm{s}), s(-r3+8D), r^{3}]$ .
Composing with (8) above, we find the isomorphism maps $C:W^{3}=2S(U^{3}-$$3rU^{2}V-4DV^{3})$ to $E’$ :
by
$O$

$Q$

$\psi$$Z’\mathrm{Y}\prime^{2}=X^{\prime 3}-27DZ’3$$X’=3(2sV+W)(2SU-rW)$(10)$\psi$:$\mathrm{Y}’=9S(2rU(U-rV)+4DV^{2}-W2)$$Z’=(2_{S}V+W\rangle^{2}$

101

Finally, we can give an explanation for the minus sign in (4) above. This isbecause the following diagram is commutative.
$C$

(11)

descent$arrow E$$\psi\downarrow$$\downarrow\tau_{()}r,s$$E’arrow\hat{\phi}E$

Following along the left and bottom, we see that $(2r, 1, -2s)$ goes to $O\in E$ . Hencethe descent map (3) takes $(2r, 1, -2s)$ to $(r, -s)$ .

REFERENCES
[Ca]
[Li]
[Ma]
[Mo]
[Q]

[Satl][Sat2]

J.W.S.Cassels, Lectures on Elliptic Curves, Cambridge University Press, 1992.
Eric Liverance, $x^{3}+y^{3}=p$ , preprint.
K. Mahler, On the lattice points on curves of genus 1, Proc. London Math. Soc. 39(1935), 431-466.
Mordell, Diophantine Equations, Academic Press, 1969.
Jordi Quer, Corps quadmtiques de 3-mng 6 et courbes elliptiques de mng12, C.R. Acad.
Sci. Paris 305 (1987), 215-218.
Philippe Satg\’e, Une g\’en\’emlisation du calcul de Selmer, S\’eminaire de Th\’eorie des Nombres, Paris, Birkh\"auser, 1983, pp. 245-265.
Philippe Satg\’e, Groupes de Sdmer et corp s cubiques, J. Number Theory 23 (1986),

294-317.

[Sill][Si12][ST][Ve]

Joseph H. Silverman, Integer points on curves of genus 1, J. London Math. Soc. 28(1983), 1-7.
Joseph H. Silverman, The
of Elliptic Curves, Springer, 1986.
Joseph H. Silverman and John Tate, $Rat\dot{i}onal$ Points on Elliptic Curves, Springer, 1992.
J. Velu, Isog\’enies entre courves elliptiques, C. R. Acad. Sc. Paris (1971), 238-241.
$A\dot{n}thmet_{\dot{i}c}$