Re: ECC Result Verification
From: Cristiano (cristiano.pi_at_NSquipo.it)
Date: 11/02/04
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Date: Tue, 02 Nov 2004 10:32:13 GMT
flip wrote:
> can someone please verify these results?
>
> E: y^2 = x^3 + 1 x + 1 (mod 23)
>
> The set of points that satisfy E are given by:
>
> {{0,1}, {0,22}, {1,7}, {1,16}, {3,10}, {3,13}, {4,0}, {5,4}, {5,19},
> {6,4}, {6,19}, {7,11}, {7,12}, {9,7}, {9,16}, {11,3}, {11,20},
> {12,4}, {12,19}, {13,7}, {13,16}, {17,3}, {17,20}, {18,3}, {18,20},
> {19,5}, {19,18}}
Ok.
> There are a total of 28 points, so # E(F23) = 28
Yes (including O).
> The order of each point is {28, 28, 28, 28, 28, 28, 2, 7, 7, 14, 14,
> 14, 14, 28, 28, 4, 4, 14, 14, 7, 7, 7, 7, 28, 28, 28, 28}
Ok.
> The cofactor (h) of each point is {1, 1, 1, 1, 1, 1, 14, 4, 4, 2, 2,
> 2, 2, 1, 1, 7, 7, 2, 2, 4, 4, 4, 4, 1, 1, 1, 1}
This comes from #E / Ord(point), so h can be trivially calculated.
> Choose the point {0, 1} as a generator point and get:
>
> 1P = (0, 1) 11P = (1, 16) 21P = (11, 20)
> 2P = (6, 19) 12P = (17, 20) 22P = (7, 12)
> 3P = (3, 13) 13P = (9, 16) 23P = (18, 20)
> 4P = (13, 16) 14P = (4, 0) 24P = (13, 7)
> 5P = (18, 3) 15P = (9, 7) 25P = (3, 10)
> 6P = (7, 11) 16P = (17, 3) 26P = (6, 4)
> 7P = (11, 3) 17P = (1, 7) 27P = (0, 22)
> 8P = (5, 19) 18P = (12, 19) 28P = (0, 1) = 1P (point at
infinity)
> 9P = (19, 18) 19P = (19, 5)
> 10P = (12, 4) 20P = (5, 4)
I don't understand 28P = (0, 1) = 1P (point at infinity); 28P = O (point at
infinity), but O != P (or 1P).
All the other points are ok.
> Lastly: 17*(19, 18) and 9*(1, 7) = (9, 16) (this did not match the
> book, which I think might be wrong).
I get 17*(19, 18) = (9,7) and 9*(1, 7) = (9, 16).
Cristiano
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