2
$\begingroup$

I'm trying to show that the rank of the following elliptic curve

$$ \mathscr{E}: y^2=x(x^2-25)$$

is 1. Since it has a rational 2-torsion point at $(0,0)$, by considering the dual curve I've been able to show that the rank is at most 1. However, to show that it is one I'm trying to find a rational point with non-integral entries (which would hence not be a torsion point, implying the rank of $\mathscr{E}$ is at least 1).

The hint we are given is to consider $x \in {\mathbb{Q}^*}^2$. Considering a potential solution $x=\frac{a^2}{b^2}$ with $\gcd(a,b)=1$, then we are reduced to solving

$$ (\dfrac{yb^3}{a})^2 = a^4-25b^4 $$

The right hand side is an integer, hence so is the left hand side, and must be some integer $n^2$. So

$$ a^4-25b^4=n^2$$

for some integers $a,b,n$ with $a$ and $b$ coprime. This is the homogenous weight space equation for the curve $\mathscr{E}$ for divisor $1$ of $-25$. I'm trying to see if there is some kind of method by descent which I can use to construct a solution, but it's proving to be a challenge. Can anyone offer a helpful hint?

$\endgroup$
  • $\begingroup$ By the way, if $n$ is a congruent number, then $x(x-n)(x+n)=y^2$ has positive rank. It is conjectured that all $n=8m+a$ for $a=5,6,7$ are congruent numbers. $\endgroup$ – Tito Piezas III Jan 9 '18 at 16:41
2
$\begingroup$

If instead of looking at $x=\frac{a^2}{b^2}$ you look at $x=-\frac{a^2}{b^2}$ you end up with $25b^4-a^4=n^2$ which has as an easy solution $a=2,b=1$. In fact the point $P=(-4,6)$ lies on the curve, and you can easily check that $2P$ has non-integral coordinates.

$\endgroup$
  • $\begingroup$ Ah! That makes a lot of sense - thank you! I notice that's also the form of equation you get if you assume that n is divisible by 5. $\endgroup$ – Aaron May 20 '15 at 11:28
  • $\begingroup$ In fact, your solution motivated me with the original idea. So we get the point $(a,b,n)=(5,2,15)$ on $ a^4-25b^4=n^2$. So we then get a point $(x,y)=(25/4, 75/8)$ on the curve. Also, how did you spot that the point $(-4,6)$ lies on the curve? $\endgroup$ – Aaron May 20 '15 at 12:16
  • $\begingroup$ substituting $a=2$ and $b=1$ in $x=-a^2/b^2$ $\endgroup$ – Ferra May 20 '15 at 12:17
  • $\begingroup$ Oh yeah, of course - I just realised that after I wrote it. I can't calculate it seems! $\endgroup$ – Aaron May 20 '15 at 12:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.