-2
$\begingroup$

are all elements of ZpxZp in ECC (elliptic curve) definite over Zp ?

otherwise: assume G a base point of ECC and n the order of G.

why n is equal or nother to p*p ? (p a prime number).

(Think to a Bitcoin Curve: y^2 = x^3 + 7).

$\endgroup$
1
$\begingroup$

All points have two coordinates. Both coordinates lies in $\mathbb{Z}_p$ (I'm assuming a curve defined over $GF(p)$).

But not all elements of $\mathbb{Z}_p \times \mathbb{Z}_p$ are valid points for an elliptic curve. Only those which verifies the curve's equation.

So if you select all elements of $\mathbb{Z}_p \times \mathbb{Z}_p$ and call the first $x$ and the second $y$, and when you put them in (for example) $y^2 = x^3 + 7$ if you obtain an equality, then your $x$ and $y$ form a valid point belonging to your curve.

If you count them (and add the point at infinity, which is a special point who can't be represented as $(x,y)$), their number will be the cardinality of the curve, which is a number related to your $p$ through Hasse's theorem but can be smaller or bigger.

| improve this answer | |
$\endgroup$
  • $\begingroup$ True. I only make a mistake in the equation that I have updated to the right one: y^2 = x^3 + 7 $\endgroup$ – Benwest Sep 1 '17 at 8:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.