# Computing fractions Weierstrass curves and DLP problem

I am preparing for a crypto exam by making an old practise exam. I got stuck on the following assignment.
I got this weierstrass curve
The curve $y^2 = x^3$ is not an elliptic curve over $F_{71}$ but the set of points $\{ (x,y)|x,y\in F_{71}^*, y^2 = x^3 \} \cup \{P_\infty \}$ Forms a group under the addition and doubling laws on (short) Weierstrass curves.

My first assignment was to compute for the point P = (1,1). 2P,3P,4P and 8P
This was fairly easy and I got the following results 2P = (18,9), 3P = (8,50), 4P = (40,10)
and 8P = (10,19).

Question b asks me the following: Compute the fractions x/y for 2P,3P,4P and 8P.
However, I have no idea what they mean by this. I searched my notes and wikipedia but can't find anything about fractions of x/y or what I gain by simply dividing the x point or y point.

Question c is the following: Compute the discrete logarithm of (6,43) with base (1,1). Make sure to justify your approach.
This is a clear question but I can't seem to find any good algorithm specified to tackle this approach.

An answer would be great! I'm not asking you to do my homework, if you could at least point out where to find the material to tackle these problems I would be grateful.

EDIT: So I naively just did what assignment b told me and got the following result:
$\frac{18}{9} = 2$
$\frac{8}{50} = 3$
$\frac{40}{10} = 4$
$\frac{10}{19} = 8$
This can't be a weird coincidence and makes question C fairly easy. But I have no clue why I should get this result..

The reason why you are getting this result is that if $E/K$ is a curve given by a singular Weierstrass equation, and $E$ has a cusp at $(0,0)$, with "tangent" line $y=0$ at $(0,0)$, then $E_{\text{ns}}(K)\cong K$ are isomorphic as abelian groups, where $E_\text{ns}(K)$ is the group formed by non-singular points on $E$ (i.e., $E(K)$ except $(0,0)$). The isomorphism is precisely given by $$\psi:E_{\text{ns}}(K) \to K,\ \text{ with } \psi((x,y))= \frac{x}{y},$$ and therefore $\psi(n\cdot(x,y))=n\cdot \frac{x}{y}$. In your case $x=y=1$, and so $\psi(n\cdot (1,1))=n\cdot 1/1=n$.