Calculating Square Roots of an Elliptic Curve

[SneakyG]

So there are four square roots for an elliptic curve represented by an equation something like this: y^2 = x^3 + x + 6 (mod 5)

How would one go about calculating these?

 

[DonAntonio]

So there are four square roots for an elliptic curve represented by an equation something like this: y^2 = x^3 + x + 6 (mod 5)

How would one go about calculating these?

To begin with, why not write the equation in modulo 5?
y^2=x^3+x+1

Let's now check the cubes and squares modulo 5:

0^2=0\,\,,\,1^2=1\,\,,\,2^2=4\,\,,\,3^2=4\,\,,\,4^2=1
0^3=0\,\,,\,1^3=1\,\,,\,2^3=3\,\,,\,3^3=2\,\,,\,4^3=4

We get at once the solutions
(0,1)\,\,,\,(0,4)\,\,,\,(2,1)\,\,,\,(2,4)\,\,,\,(3,1)\,\,,\,(3,4)\,\,,\,(4,2)\,\,,\,(4,3)

DonAntonio

 

[SneakyG]

Thanks. How do you calculate the orders?

 

[DonAntonio]

Thanks. How do you calculate the orders?

Apply the group law to the points...you know it, right? Otherwise it'll be impossible for you to understand what's

going on. You can read this in Silverman's "The Arithmetic of Elliptic Curves", for example. Let's do one of them, say:

(0,1)+(0,1)=(4,2)\,\,,\,\,(0,1)+(4,2)=(1,0)=0=\,\,\text{the group's zero}\,

So the element \,(0,1)\in\Bbb E(\Bbb F_5)\, has order \,3\, ...

DonAntonio