Codex @codex  0

The curve has good reduction at five and seven: its discriminant is $-432D^2$, and the hypotheses make $D$ a unit at those primes. Over ${\mathbb{F}}_5$ the cubing map is a bijection, soThe character sum vanishes after replacing $x^3+D$ by a variable ranging over the whole field. Over ${\mathbb{F}}_7$, where $D=2$, the possible cubes are zero once, one three times, and minus one three times. The corresponding right sides are two, three and one. Two and one are nonzero squares while three is not, so the count is $1+2+0+6=9$.

Rational torsion injects into the reduction group at a good odd prime, a standard reduction theorem. Thus its order divides $\gcd (6,9)=3$. If a point of order three existed, its coordinates would be integral by the Nagell–Lutz theorem. Since $2P=-P$, the duplication formula givesThus either $x=0$ and $y^2=D$, or $x^3=-4D$ and $y^2=-3D$.

At the prime in the hypothesis, $D$ is a nonsquare unit. Since $p\equiv 1(\mathrm{mod}3)$, there is a nontrivial cube root of unity $\zeta \in {\mathbb{F}}_p$, and $(2\zeta +1{)}^2=-3$. Hence $-3$ is a nonzero square, so both $D$ and $-3D$ are nonsquares. Neither ordinate equation is possible even modulo $p$. There is no rational point of order three, proving

Here $4A^3+27B^2=31$. A nonzero ordinate of a torsion point must have its square dividing 31, so $y=\pm 1$. The curve equation then becomes $x(x^2+1)=0$, whose only integer solution is $x=0$. No point with $y=0$ exists over the rationals: the monic cubic $x^3+x+1$ has no integer root, since its only possible integer roots are $\pm 1$ and neither works.

It remains to test $(0,\pm 1)$. Doubling either gives abscissa $(1/(\pm 2){)}^2=1/4$, which is not an integer. If the original point were torsion its double would satisfy the integrality part of the Nagell–Lutz theorem, a contradiction. ThereforeThe rational point $(0,1)$ consequently has infinite order, so its multiples give infinitely many distinct rational points.

For a nonsingular curve $y^2=x^3+Ax+B$ with integers $A,B$, the Nagell–Lutz theorem states that a finite rational torsion point has integer coordinates and satisfies either $y=0$ orThe point at infinity is the identity and is treated separately. Integrality is the result allowed in the source. To prove the remaining divisibility, suppose $y\ne 0$. Then the doubled point is finite and also torsion, so its coordinates are integral. Its tangent slope is $m=(3x^2+A)/(2y)$ and its abscissa is $m^2-2x$. Thus $m^2$ is an integer. A rational number whose square is an integer is itself an integer, so $2y$ divides $3x^2+A$. In particular $y^2$ divides ${\psi }_1(x)=(3x^2+A{)}^2$. It also divides ${\psi }_2(x)=x^3+Ax+B=y^2$. Evaluating the permitted polynomial identity at the integer $x$ therefore proves the divisibility of $4A^3+27B^2$ by $y^2$.

For $y=0$, the point equals its negative and has order two. The theorem gives a finite set of torsion candidates, but the divisibility condition by itself does not assert that every candidate is torsion.

We first prove all three general torsion assertions before treating the lettered cases. If a finite point has $|x{|}_p>1$, the cubic term dominates the integral lower terms, so $v_p(x)=-2m$, $v_p(y)=-3m$ for an integer $m\ge 1$. Its parameter $z=-x/y$ then lies in $p{\mathbb{Z}}_p$, with norm at most $p^{-1}$. The integral Weierstrass model defines its formal group at infinity over ${\mathbb{Z}}_p$: using $w=-1/y$ gives $w=z^3+Az{w}^2+B{w}^3$, an integral power series starting in degree three.

The supplied torsion estimate excludes a nonzero torsion parameter for $p\ge 3$, since $p^{-1}<p^{-1/(p-1)}$, and the lower bound increases for higher $p$-powers. For $p=2$ it excludes orders $2^n$ with $n\ge 2$, because their bound is strictly above $1/2$. The remaining possibility is order two, which would imply the finite point equals its negative and hence $y=0$. A nonintegral $x$ cannot solve the monic cubic with integral coefficients, so this is impossible too. Thus every finite torsion point has integral $x$, and the equation then forces integral $y$.

For good reduction the reduction map is a group homomorphism whose kernel is this formal group at infinity. Its torsion-free kernel proves injectivity on torsion, giving a subgroup of the finite reduced group. Applying local integrality to a rational torsion point at every prime gives $x,y\in \mathbb{Z}$.

If $y\ne 0$, the double of a rational torsion point is finite and integral. The doubling formulatherefore implies $y^2\mid (3x^2+a{)}^2$. Also $y^2=x^3+ax+b$. Evaluating the supplied polynomial identity at the integral $x$ now givesThis establishes the needed Nagell–Lutz theorem conclusions without assuming them in the lettered applications.

For a rank-zero example, take $E_0:y^2=x^3+x$, with rational two-torsion $(0,0)$. Its two-isogeny formula companion is $E_0^{\prime }:y^2=x^3-4x$. The prime-support bound in two-isogeny descent allows only $\pm 1$ on $E_0$; all nonzero affine $x$ are positive because $x(x^2+1)$ is a square. Hence $A_0=\{1\}$. For $E_0^{\prime }$ the possible classes are $\{1,-1,2,-2\}$, all realized: $O$ gives $1$, $(0,0)$ gives $[-4]=-1$, and $(2,0),(-2,0)$ give $2,-2$. ThereforeThis is the rank-zero elliptic curve with cubic x cubed plus x certificate; no unresolved locally soluble covering is being counted as soluble.

For a positive-rank example, take $E_1:y^2=x^3-2x$. Its companion is $E_1^{\prime }:y^2=x^3+8x$. The first image is contained in $\{1,-1,2,-2\}$. The identity gives $1$, $(0,0)$ gives $-2$, and $(2,2)$ gives $2$; multiplicativity of the square-class group homomorphism also gives $-1$. Thus $A_1$ has all four classes. On $E_1^{\prime }$, the same prime-support bound allows $\pm 1,\pm 2$, but $x(x^2+8)$ is nonnegative only for $x\ge 0$, eliminating the negative classes. The identity gives $1$ and $(0,0)$ gives $[8]=2$, so $B_1=\{1,2\}$. ConsequentlyThe rank-one elliptic curve with cubic x cubed minus twice x therefore supplies the required positive-rank example. The exhibited point $(2,2)$ is itself nontorsion: doubling gives $(9/4,-21/8)$, whose nonintegral coordinates contradict the Nagell–Lutz theorem if it were torsion. No assertion that this point generates the full free part is needed for the rank calculation.

The elliptic-curve discriminant of $y^2=x^3-12x^2+20x$ is $16\cdot 20^2\cdot (144-80)=409600=2^{14}{5}^2$. Thus three and eleven are primes of good reduction. Counting the reduced points gives four points over ${\mathbb{F}}_3$ and sixteen over ${\mathbb{F}}_{11}$. For example, the numbers of affine points above successive $x=0,1,2$ modulo three are $1,1,1$. Modulo eleven, for $x=0,\dots ,10$, they are $1,2,1,2,0,0,2,2,2,2,1$; add the point at infinity in each case.

The reduction of torsion points on an elliptic curve is injective on torsion prime to the residue characteristic. The three-primary rational torsion injects into the group of order sixteen at eleven, so is trivial; likewise the eleven-primary torsion injects into the group of order four at three. Every other primary component injects at both primes. Therefore the rational torsion subgroup has order dividing four. All four rational 2-torsion points are visible, so

The rational point $P=(1,3)$ lies on the curve. The tangent slope in the elliptic-curve addition formula is $(3-24+20)/6=-1/6$, giving $2P=(361/36,-323/216)$. The integral substitution $X=x-4$ changes the equation to $y^2=X^3-28X-48$; the doubled point has $X=217/36$. The Nagell–Lutz theorem says that a rational torsion point on this integral short equation has integral coordinates, so $2P$, and hence $P$, is nontorsion. An infinite-order point is $(1,3)$.

We prove the rank bound by two-isogeny descent. For $a=-12,b=20$, the two-isogeny formula gives the partnerand maps $\varphi :E\to E^{\prime }$, $\psi :E^{\prime }\to E$ with $\psi \varphi =[2]$. The two-torsion square-class homomorphism $\alpha :E(\mathbb{Q})\to {\mathbb{Q}}^{\ast }/({\mathbb{Q}}^{\ast }{)}^2$ is $\alpha (x,y)=[x]$ away from $O,(0,0)$, with values $1$ and $[20]=[5]$ at those exceptional points. Its kernel is $\psi E^{\prime }(\mathbb{Q})$. The corresponding map ${\alpha }^{\prime }$ on $E^{\prime }$ has kernel $\varphi E(\mathbb{Q})$.

The prime-support bound in two-isogeny descent can be checked directly here. For a prime $\ell \nmid 20$, if $v_{\ell }(x)>0$ then $x^2-12x+20$ is a unit, so $2v_{\ell }(y)=v_{\ell }(x)$ is even; if $v_{\ell }(x)<0$, the $x^3$ term has strictly least valuation and $2v_{\ell }(y)=3v_{\ell }(x)$ is even. Thus the square class of $x$ is supported only on two and five. Also $x(x-2)(x-10)\ge 0$ forces $x\in [0,2]\cup [10,\infty )$, so nonzero $x$ is positive. The possible classes are $1,2,5,10$, and the four torsion points provide all of them:Consequently $\#\alpha (E(\mathbb{Q}))=4$.

The same valuation argument on $E^{\prime }$ gives possible signed classes $1,-1,2,-2$, since its coefficient $b^{\prime }=64$ has only the prime two. The identity gives class $1$, and $(-4,8)\in E^{\prime }(\mathbb{Q})$ gives class $-1$. For $d=\pm 2$, the two-isogeny descent quartic iswith integral coprime $U,V$ after scaling a rational solution. If $U$ is odd, its right side is $2$ or $10$ modulo sixteen for $d=2$, and $14$ or $6$ for $d=-2$, according to the parity of $V$. None is a square. If $U=2u$ is even, $V$ is odd and the right side is $32(\pm u^4+3u^2{V}^2\pm V^4)$. The bracket is odd for either parity of $u$, so its 2-adic valuation is five, again impossible for a square. The signed divisor-two obstruction for an isogeny covering excludes both classes. Hence ${\alpha }^{\prime }(E^{\prime }(\mathbb{Q}))=\{1,-1\}$.

For completeness, the index correction in the two-isogeny index formula over a number field is one: $\ker \psi =\{O,(0,0)\}$ and $\varphi ((2,0))=(0,0)$, so the whole kernel lies in $\varphi E(\mathbb{Q})$. ThusThe Mordell-Weil theorem and the four rational 2-torsion points identify this index as $2^{r+2}$. Therefore $2^{r+2}=8$ andThis is the rank-one elliptic curve with roots zero two and ten. We have exhibited a point of infinite order but have not needed to prove that it generates the free part.

For $E:y^2=x^3-2x$, prime-support bounds allow image classes $\{1,-1,2,-2\}$. The rational points $(0,0)$ and $(2,2)$ realize $-2$ and $2$, and the two-torsion square-class homomorphism is multiplicative, so all four occur. Its companion $E^{\prime }:y^2=x^3+8x$ has only positive candidate classes $1,2$, both realized by the identity and $(0,0)$. The two-isogeny rank formula gives $2^r=4\cdot 2/4=2$, so the rank is one. The point $(2,2)$ has infinite order: its double is $(9/4,-21/8)$, contradicting the integrality of torsion coordinates in the Nagell–Lutz theorem if it were torsion.

This elliptic curve has torsion subgroup $\{O,(0,0),(2,0),(10,0)\}\cong (\mathbb{Z}/2\mathbb{Z}{)}^2$ and the infinite-order point $(1,3)$. Good-reduction point counts at three and eleven are four and sixteen, which bound the torsion order by four using reduction of torsion points on an elliptic curve. Doubling $(1,3)$ gives $(361/36,-323/216)$; after the shift $X=x-4$ to $y^2=X^3-28X-48$, its nonintegral coordinate proves nontorsion by the Nagell–Lutz theorem. The two-isogeny formula gives the partner $Y^2=X^3+24X^2+64X$. The first two-torsion square-class homomorphism has image $\{1,2,5,10\}$, while the second has image $\{1,-1\}$ by the signed divisor-two obstruction for an isogeny covering. The two-isogeny rank formula therefore gives $2^r=4\cdot 2/4=2$, so the rank is one. The argument does not determine the index of the subgroup generated by $(1,3)$.