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I am currently revising for an exam and need some help with a question. Below is an example from my notes which I am trying to understand. I can fill out the table just fine, but I can't figure out how to find the points from it. If someone could run through getting each point I would be grateful.

Example 1: Let p = 5 and y2 = x3-3x^2+3x. Draw up a table of values of 
x^3-3x^2+3x.

x  x^2   x^3   -3x^2   3x   x^3-3x^2+3x
0   0     0      0      0        0
1   1     1      2      3        1
2   4     3      3      1        2
3   4     2      3      4        4
4   1     4      2      2        3

Now mod p, 0^2=0, 1^2=1, 2^2=4, 3^2=4 and 4^2=1. 

(please note that the equal signs in the above line are congruence signs in notes, I don't know how to put them in here)

Thus we can list the points of our elliptic curve mod p.

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

up vote 1 down vote accepted

You know that for say $x=1$, $y^2 = 1$. The solutions to this equation (mod 5) are $y = 4$ and $y=1$. The same goes for the other points. To repeat,in more generality, the only thing you need to do is to figure out what y has the property that $y^2= a^3-3a^2+3a$, where $a \in \mathbb{Z}/5\mathbb{Z}$. Say that $b \in \mathbb{Z}/5\mathbb{Z}$ has the property that $b^2 = a^3-3a^2+3a$. Then $(a,b)$ is a point on the curve.

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Thank you for the response. I don't quite grasp what you are saying, so I have a question or two if you don't mind? How does the output of the table feed into the equations? For example, 0, 1, 4, 4, 1 is the list of numbers modulo 5. Do I test each one of them as x and use the result as my y? Or am I feeding in 0-4 as my x variable? –  David Ryan Dec 20 '12 at 12:27
    
I do not mind. You test with 0,1,2,3,4 for x. This will gives rise to a value of $x^3-3x^2+3x$. This is $y^2$. To get the value of y, you must then find what y satisfies this equation. –  Dedalus Dec 20 '12 at 13:31
    
Thanks very much, i get it now. Merry Christmas! –  David Ryan Dec 20 '12 at 13:55

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