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Cracking parameter b of the elliptic curve

Today, 24 November 2004, I found the answer to the question why #E0 + #E1 = 2(q + 1).  It is actually quite trivial.  The reason is simply this:

The number of points on a given elliptic curve is equal to

#E = 2 + 2$\kappa$

where $\kappa$ is the number of different non-zero x coordinates for which there is a solution (and thus exactly two solutions) to the elliptic curve equation E: x3 + ax2 + b = y2 + xy.

The first 2 takes the point at infinity and the point with x is zero (0, $\sqrt{b}$) into account.

After dividing the elliptic curve equation by x2, we see that there will be a solution for a non-zero value of x when x + a + b/x2 = (y/x)2 + y/x.  And because of theorem 5, this only has a solution when Tr(x + a + b/x2) = 0.

In other words, when the trace of a is changed from 0 into 1, then there will be solutions for exactly the other non-zero values of x!

Therefore, with #E0 = 2 + 2$\kappa_0$ and #E1 = 2 + 2$\kappa_1$ as defined in the chapter "Cracking parameter a of the elliptic curve", we can conclude that $\kappa_0$ + $\kappa_1$ equals the total number of possible non-zero values of the field: q - 1.  And thus #E0 + #E1 = 4 + 2(q - 1) = 2(q + 1).

The following might be a very important result, since I never saw it anywhere in literature yet (please contact me if you know whether or not this is indeed new):

The order of an elliptic curve over $\mathbb{F}_{2^m}$ is equal to (assuming Tr(a) = 0, otherwise it is 2(q + 1) minus this) 2 plus 2 times the number of values of x for which Tr(x) = Tr(b/x2).

If its new, then I hereby claim it ;).

<To Be Continued>

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