Exercise 5.7.16 ($y^2 = x^3 + 6x^2 + 2x$ contains no rational point with $x<0$)

Proof. If $b=0$, then $p(x)=x^3+a{x}^2=x^2(x+a)$ has $0$ as a double root, so the curve ${𝒞}_f(ℝ)$ has a singularity at $(0,0)$. Therefore ${𝒞}_f(ℝ)$ is not an elliptic curve, contrary to the hypothesis. So $b\ne 0$.

Suppose that $(x_1,y_1)=(u_1∕w_1^2,v_1∕w_1^3)$ is a rational point on this curve, with $u_1\wedge w_1=1$. As in Problem 13, with $c=0$, since $P_1=(x_1,y_1)=(u_1∕w_1^2,v_1∕w_1^3)\in {𝒞}_f(\mathbb{Q})$, the equation of $f$ gives

that is

Let $d=u_1\wedge b$ be the g.c.d. of $u_1$ and $b$. Since $b\ne 0$, we have $d>0$. There are integers $U,B$ such that

Substituting $u_1$ and $b$ in (1), we obtain

thus $d^2\mid v_1^2$, therefore $d\mid v_1$: we write $v_1=\mathit{dV}$ where $V$ is an integer, so

If we simplify by $d^2$,

Moreover,

because $U\wedge B=1$, and $U\mid u_1$, where $u_1\wedge w_1=1$, thus $U\wedge w_1$: this shows that $U\wedge (B{w}_1^4)=1$.

Put $W=d{U}^2+a{w}_1^{2}U+B{w}_1^4$. Then $V^2=\mathit{UW}$, and $U\wedge W=1$, therefore $U=\pm s^2$ for some integer $s$ (by Lemma 5.4). We have shown that there exist integers $d$ and $s$ such that $d>0$, $d\mid b$, and $u_1=\pm d{s}^2$.

Consider the particular case of the curve $𝒞:y^2=x^3+6x^2+2x$.

If $(x_1,y_1)=(u_1∕w_1^2,v_1∕w_1^3)$ is a rational point on $𝒞$, where $(x_1,y_1)\ne (0,0)$, then $u_1=\pm d{s}^2$, where $d>0$ and $d\mid b=2$, so $d=1$ or $d=2$. This gives

for some integer $s\in {\mathbb{N}}^{\text{∗}}$. Moreover, by (1),

• Suppose that $u_1=-s^2$. Then by (2),

  $$\begin{array}{lll}\hfill v_1^2=-s^6+6s^4{w}_1^2-2s^2{w}_1^4.& & \hfill \text{(3)}\end{array}$$

  This shows that $s^2\mid v_1^2$. Therefore $s\mid v_1$, so $v_1=\mathit{sV}$ for some integer $V$, and simplifying by $s^2$, this gives

  $$\begin{array}{lll}\hfill V^2& =-s^4+6s^2{w}_1^2-2w_1^4& \hfill \text{(4)}\end{array}$$

  • If $s\equiv 1(\mathit{mod}2)$, then $s^2\equiv 1(\mathit{mod}4)$, and $w_1^2\equiv 1(\mathit{mod}4)$ or $w_1^2\equiv 0(\mathit{mod}4)$. In both cases,

    $$V^2\equiv -1(\mathit{mod}4).$$

    This congruence has no solution modulo $4$.

  • If $s\equiv 0(\mathit{mod}2)$, then $u_1=-s^2\equiv 0(\mathit{mod}4)$. Since $u_1\wedge w_1=1$ and $v_1\wedge w_1=1$, then $w_1$ is odd, and $v_1$ is even. Therefore $w_1^2\equiv 1(\mathit{mod}4)$ and $s^2\equiv 0(\mathit{mod}4)$. This gives

    $$V^2\equiv 2(\mathit{mod}4),$$

    which has no solution modulo $4$

  This shows that $u_1=-s^2$ yields no solution.

• Suppose that $u_1=-2s^2$. Then by (2),

  $$\begin{array}{lll}\hfill v_1^2=-8s^6+24s^2{w}_1^2+4s^2{w}_1^4.& & \hfill \text{(5)}\end{array}$$

  Then ${(2s)}^2\mid v_1^2$, therefore $2s\mid v_1$, so $v_1=2\mathit{sW}$ for some integer $W$. Substituting in (5), we obtain after simplification by $4s^2$

  $$\begin{array}{lll}\hfill W^2=-2s^4+6w_1^2+w_1^4.& & \hfill \text{(6)}\end{array}$$
  • If $s\equiv 1(\mathit{mod}2)$, then $s^4\equiv 1(\mathit{mod}8)$, so $W^2\equiv -2+2w_1^2+w_1^4\equiv 5$ or $6(\mathit{mod}8)$, but $W^2\equiv 0,1,$ or $4(\mathit{mod}8)$. This is impossible.

  • If $s\equiv 0(\mathit{mod}2)$, then $s^4\equiv 0(\mathit{mod}8)$, so $W^2\equiv 2w_1^2+w_1^4(\mathit{mod}8)$.

    Since $u_1=-2s^2$ is even, and $u_1\wedge w_1=1$, we know that $w_1$ is odd, thus $w_1^2\equiv 1(\mathit{mod}8)$, and $W^2\equiv 3(\mathit{mod}8)$. This is impossible.

  This shows that $u_1=-2s^2$ also gives no solution.

It remains only the possibility $u_1=s^2$ or $u_1=2s^2$. Therefore $x_1=u_1∕v_1^2\ge 0$. The elliptic curve $y^2=x^3+6x^2+2x$ contains no rational point $(x_1,y_1)$ with $x_1<0$ (there are rational points with $x_1\ge 0$, for instance $(0,0)$ or $(1,3)$). □