# Finding a rational point on $\mathscr{E} : y^2=x(x^2-25)$ to show $\text{rank}(\mathscr{E})=1$

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?

• 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. – Tito Piezas III Jan 9 '18 at 16:41

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.
• 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? – Aaron May 20 '15 at 12:16
• substituting $a=2$ and $b=1$ in $x=-a^2/b^2$ – Ferra May 20 '15 at 12:17