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Please see the attached image for a question on why the order of an elliptic curve does not consider the point at infinity in order to avoid contradicting Lagrange's Theorem.

Elliptic Curve over Finite Field specific Example

Matlab Plotting of the Elliptic Curve

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  1. It would be convenient for readers if you presented the question in text, not in a picture.
  2. Drawing curves over finite fields in MATLAB is not terribly illuminating.
  3. This is not an elliptic curve: it is $y^2=(x+4)^2(x+3)$, a rational curve with a node.
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  • $\begingroup$ Apologies for the incorrect formatting and the inclusion of a MATLAB graph (I did it to show that I was not pulling the group order number out of my head, but I understand that I could have included a table instead). Can you please elaborate on why the curve y^2=x^3 - 4x +4 is not an elliptic curve? The discriminant is non-zero. Although this is one example, I have noticed that this occurs with other Elliptic Curves. $\endgroup$
    – Harry Alli
    Jun 25, 2017 at 9:39
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    $\begingroup$ @CallumN: The field is $\mathbb{F}_{11}$ here, so the discriminant is zero. $\endgroup$
    – ccorn
    Jun 25, 2017 at 9:41
  • $\begingroup$ @CallumN It is not an elliptic curve, since its discriminant is zero. One can see this immediately since the RHS has a repeated factor. $\endgroup$ Jun 25, 2017 at 9:41
  • $\begingroup$ @ccorn and Lord Shark: So the discriminant is also calculated modulo 11? Okay, thank you. I noticed that if the discriminant is non-zero, then the curve has a prime order. However, does this mean that every point is a generator? When we include the point at infinity, the elliptic curve no longer has prime order, so going back to my original question, do we include the point at infinity? Would doing so violate Lagrange's Theorem? $\endgroup$
    – Harry Alli
    Jun 25, 2017 at 9:59
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    $\begingroup$ @CallumN Yes, you have to include the point at infinity, otherwise it won't be a group. $\endgroup$
    – themaker
    Jun 25, 2017 at 10:33

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