Rational Points on the genus 1 curve $y^2=x^4−62x^2+1$

Question

Am trying to find rational Points on the genus 1 curve $$y^2=x^4−62x^2+1$$

Or show no non trivial solutions exist.

Initially, I can't find a solution other than (x,y)= (0,1),(0,-1)

How should I proceed?

Answer 1

Oh well, I’ve never actually determined the points of an elliptic curve, so why not this once.

By the Elliptic Curve Handbook (linked in a comment), solving $v^2=u^4-62x^2+1$ is equivalent to solving $y^2=x^3-62x^2-4x+248=(x-2)(x+2)(x-62)$, by setting $x=\frac{2(v+1)}{u^2}$ and $y=\frac{4(v+1)-124u^2}{u^3}$ (the ECH has the inverse transformation as well).

The solution $(u,v)=(0,1)$ corresponds to the point at infinity, and $(u,v)=(0,-1)$ to $(x,y)=(62,0)$. We can see that the two other torsion points correspond to the one point at infinity for $v^2=u^4-62u^2+1$ (this curve is, I believe, not smooth at infinity, hence the conflation of the two points).

The discriminant of the cubic is $2^{20}\cdot 3^2 \cdot 5^2$.

It’s easy to see (using the Weierstrass equation) that at $p=7$, the elliptic curve $E$ has good reduction, so the prime-to-$7$ rational torsion injects (by Nagell-Nutz and others) in $E(\mathbb{F}_7)$, which we can check is of cardinality $8$. There is also good reduction at $p=11$ and $E(\mathbb{F}_{11})$ has cardinality $8$, so the rational torsion is either $\mathbb{Z}/(2) \oplus \mathbb{Z}/(4)$ or $\mathbb{Z}/(2) \oplus \mathbb{Z}/(2)$.

By the Elliptic Curve Handbook, we have an injective map $\Phi=(\phi_1,\phi_2,\phi_3): E(\mathbb{Q})/2E(\mathbb{Q}) \rightarrow \Sigma$ given by $\phi_1(x,y)=[x-2]$ when $x \neq 2$, $\phi_2(x,y)=[x+2]$ when $x \neq -2$, $\phi_3(x,y)=[x-62]$ when $x \neq 62$, and $\Sigma=\{(u,v,w) \in (\mathbb{Q}^{\times}/\mathbb{Q}^{\times 2})^3,\, uvw=1\}$.

In particular, we compute that $\Phi(2,0)=(-15,1,-15)$, $\Phi(-2,0)=(-1,1,-1)$, $\Phi(62,0)=(15,1,15)$.

So $\Phi$ doesn’t vanish at $(\pm 2,0)$ or $(62,0)$, so they’re not in $2E(\mathbb{Q})$ and these points are the only rational torsion points.

Now, we use the ECH, Prop. 3.6.4. Recall that the equation of the elliptic curve is $y^2=(x-2)((x-2)^2-56(x-2)-240)$.

It follows that the image of $\phi_1$ is (apart from $1$) the set of $b_1|240$ (up to square factors) such that there is a nontrivial integer solution to $n^2=b_1m^4-56m^2e^2+b_2e^4$ with $b_2=-240/b_1$.

Because of the $2$-torsion, we know that $\pm 1, \pm 15$ already work. To show that there are no other solutions, all we need to do is (using the group property) prove it for $b_1=2,3,6$.

Case 1: $b_1=2$. We consider any integral solution for $n^2=2m^4-56m^2e^2-120e^4$. We see that $8|n^2-2m^4$, which can only hold if $n$ is divisible by $4$ and $m$ is even. But in this case, $2m^4, n^2$ and $56m^2e^2$ are divisible by $16$, so $16|120e^4$, so we also know that $2|e$. By infinite descent there is no nontrivial solution.

Case 2: $b_1=6$. The argument is identical.

Case 3: $b_1=3$: our equation is $n^2=3m^4-56m^2e^2-80e^4$, and amenable to the same analysis with the same conclusion.

So the image of $\phi_1$ is exactly $\{\pm 15, \pm 1\}$.

Next, write the elliptic curve as $y^2=(x+2)((x+2)^2-68(x+2)+256)$. To show that $\phi_2=1$, by the same proposition, we only need to show that $n^2=b_1m^4-68m^2e^2+256e^4/b_1$ has no nontrivial integral solutions for $b_1=-1,-2,2$. When $b_1<0$, $256/b_1<0$ so the RHS is nonnegative iff $m=e=0=n$, QED.

So all that remains to do is the case $b_2=2$, ie we want to show that there are no integer solutions to $n^2=2m^4-68m^2e^2+128e^4$ with $m$ and $e$ coprime. Indeed, this equation reduces mod $3$ to $n^2=2(m^2+e^2)^2$, which implies that $3$ divides $m$ and $e$, a contradiction.

Because $\phi_2$ is trivial, the value of $\phi_1$ determines the value of $\Phi$, so $\phi_1$ is injective. Thus the image of $\phi_1$ has four elements: so $E(\mathbb{Q})/2E(\mathbb{Q})$ has cardinality at most four, and therefore $E[2](\mathbb{Q}) \rightarrow E(\mathbb{Q})/2E(\mathbb{Q})$ is an isomorphism. It follows by Mordell-Weil that $E(\mathbb{Q})$ is exactly the $2$-torsion.

By the discussion at the beginning, it shows that $(0,\pm 1)$ are the only solutions to the original equation.

Answer 2

Partial answer.-There are not integer solutions other than the aforementioned $(x,y)=(0,\pm1)$.

We have equivalently $y^2=(x^2-31)^2-960=2^6\cdot3\cdot5$ Put $x^2-31=z$ so one has $$z^2-y^2=(z+y)(z-y)=960$$ The integer $960$ has $(6+1)(1+1)(1+1)=28$ divisors which gives after calculation the possible positive values $z=241,122,83,64,53,46,38,34,32,31$. However $x^2-31=z$ has no integer solution for these values of $z$ excepting the trivial ones $(x,y)=(0,\pm1)$ with $z=31$.

For rational solutions, it is a difficult problem concerning elliptic curves if what have said by the O.P. about the genus $1$ were correct.