Does there exist a rational point on the elliptic curve: $y^2=x^3+6x^2+x$ ? If yes, how to find one? (relations to the 'rational distance problem') [closed]

Question

The elliptic curve $y^2=x^3+6x^2+x$ is associated with the Rational Distance problem, which asks whether there exists a point in the plane, that is at rational distances from the four vertices of the square (with rational side). This particular elliptic curve encodes the existence of rational points on the line bisecting any one side of the square (preferably a unit square).

So, as a part of the problem, it requires us to find two right triangles, with length of perpendiculars, $1/2$ and the sum or difference of the base lengths to be unity. The sides of the right triangle can be generated via Pythagorean triples. So, there are two triangles of sides, $\frac{p^2-1} {4p},1/2, \frac{p^2+1} {4p}$ and $\frac{q^2-1} {4q},1/2, \frac{q^2+1} {4q}$, for some parameters $p$ and $q$. As, the sum or difference in base lengths are 1, so, $$\frac{p^2-1} {4p} \pm \frac{q^2-1} {4q}=1$$ Taking positive sign for interior points, and putting $q=np$ for some $n\in Q$, we obtain, $$p=\frac{2}{n+1}\pm \frac{\sqrt{n^3+6n^2+n}} {n(n+1)}$$ Now replacing $n$ with $x$ and setting the term in the square root equal to $y^2$, we get the elliptic equation,$$y^2=x^3+6x^2+x$$ So if we can find a rational point on this particular elliptic curve, we can easily get a point in the plane that is at rational distance from all the vertices of the square.

Answer 1

Here is the $2$-descent, following Silverman, The Arithmetic of Elliptic Curves, Proposition X.4.9 and Example X.4.10.

Let $E: y^2 = x^3 + 6x^2 + x$ and $E': Y^2= X^3-12X^2+32X$ and $\phi: E \to E', (x,y) \mapsto(y^2/x^2,y(1-x^2)/x^2)$. This has the $\mathbf{Q}$-rational $2$-torsion point $(0,0)$. Then in Silverman's notation, one has $a = 6, b = 1$. The discriminant is $\Delta_E = 2^9$, so $S = \{2,\infty\}$ and $\mathbf{Q}(2,S) = \{\pm1,\pm2\}$.

For $d \in \mathbf{Q}^\times$, one has the principal homogeneous space $C_d: dw^2 = d^2- 12dz^2+32z^4$. The Selmer group is $\mathrm{Sel}^{(\phi)}(E/\mathbf{Q}) = \{d \in \mathbf{Q}(2,S):C_d(\mathbf{Q}_v) \neq \emptyset\,\,\forall v \in S\}$.

If $d < 0$, clearly $C_d(\mathbf{R}) = \emptyset$. (For $d = 1$, $C_1: w^2 = 1 - 12z^2 + 32z^4$ has the $\mathbf{Q}$-rational point $(w,z) = (0,\frac{1}{2})$, so $1 \in \mathrm{Sel}^{(\phi)}(E/\mathbf{Q})$.) For $d = 2$, $C_1: 2w^2 = 4 - 24z^2 + 32z^4$ has the $\mathbf{Q}$-rational point $(w,z) = (0,\frac{1}{2})$, so $2 \in \mathrm{Sel}^{(\phi)}(E/\mathbf{Q})$.

Hence $\mathrm{Sel}^{(\phi)}(E/\mathbf{Q}) \cong \mathbf{Z}/2$. One has the exact sequence $0 \to E'(\mathbf{Q})/\phi(E(\mathbf{Q})) \to \mathrm{Sel}^{(\phi)}(E/\mathbf{Q}) \to \mathrm{III}(E/\mathbf{Q})[\phi] \to 0$. $E'$ has the $\mathbf{Q}$-rational $2$-torsion point $(0,0)$, so $\mathrm{III}(E/\mathbf{Q})[\phi] = 0$ and $E'(\mathbf{Q})/\phi(E(\mathbf{Q})) \cong \mathbf{Z}/2$ generated by $(0,0)$. Similarly, $\mathrm{III}(E'/\mathbf{Q})[\hat{\phi}] = 0$. One gets $E(\mathbf{Q})/2 \cong \mathbf{Z}/2$ and that the rank is $0$ and the Mordell-Weil groups are $E(\mathbf{Q}) \cong \mathbf{Z}/4$ and $E'(\mathbf{Q}) \cong (\mathbf{Z}/2)^2$.

(In fact, Magma calculations show that the Tate-Shafarevich group is trivial.)

Answer 2

There's also the LMFDB ($L$-Series and Modular Forms Database), where you can look up things like this. Making the change of variables $x\to x-2$ gives the Weierstrass equation $$ Y^2 = X^3-11X+14. $$ This is Elliptic Curve 32.a2 in the database: http://www.lmfdb.org/EllipticCurve/Q/32/a/2

There's lots of information compiled there, including the fact that this curve has complex multiplication, $\text{End}(E)\cong\mathbb Z[\sqrt{-4}]$. And, of course, it says that $E(\mathbb Q)\cong\mathbb Z/4\mathbb Z$ with generator $(1,2)$, as Timo has calculated for you.