If I have $E: y^2 = 2x^3 + 3$, and I want to compute all the points over $\mathbb{F}_5$

Do I simply just plug in 0-4 for y, and 0-4 for $2x^3 +3$ and then all the ones that are equal are the only points? So in this example

$$(1,0), (2,2), (2,3), (4,1), (4,4)$$

How do I find the points at infinity?


  • $\begingroup$ It's one way to do it. An elliptic curve in Weierstrass form has one point at infinity. $\endgroup$ – Lord Shark the Unknown Mar 8 '18 at 20:06

By definition, the group $E(\mathbb{F}_5)$ is given by $$ E(\mathbb{F}_5)=\{(x,y)\mid y^2=2x^3+3\}\cup \{\mathcal{O}\}, $$ where $\mathcal{O}$ is the point "at infinity". This extra point satisfies $P+\mathcal{O}=\mathcal{O}+P=P$ for the point addition on $E$. One can check the size of such groups by the Hasse-inequality $$ |p+1-\# E(\mathbb{F}_p)|\le 2\sqrt{p}. $$


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