I am attempting to understand ElGamal Encryption on Elliptic Curves (beginnning of this paper)

I have an elliptic curve $E\left( \mathbb{F}_{11} \right)$ defined by $y^2 = x^3 + 2x + 3$

I need to choose a point $P$ on $E\left(\mathbb{F}_{11}\right)$ of prime order $n$

I haven't been able to find anything to help me understand what is meant by 'prime order $n$'. Can someone point me in the direction of anything useful or explain to me how to find such a point $P$ please?

I am a computer scientist, so assume limited knowledge of any complex maths/theorems - all my knowledge on elliptic curves is self-taught through Googling...

  • 1
    $\begingroup$ $E(\mathbb{F}_{11})$ is a finite abelian group whose elements are the points of the curve. $P$ has order $n$ means that $n$ is the least integer such that $\underbrace{P\oplus\ldots\oplus P}_n = 0$ where $0$ is the identity of the group. You have an example there $\endgroup$
    – reuns
    Feb 13, 2017 at 12:01
  • $\begingroup$ Thanks @user1952009, this makes sense - so saying it has prime order $n$ just means $n$ has to be prime in your definition, right? $\endgroup$
    – lioness99a
    Feb 13, 2017 at 12:37

2 Answers 2


You were already given a nice answer in comments, but lets add a little detail.

We find all of the points on the curve (Note that $O$ is the point at infinity)

$$(x, y) = O, (0, 5), (0, 6), (2, 2), (2, 9), (3, 5), (3, 6), (4, 3), (4, 8), (6, 0), (8, 5), (8, 6), (10, 0)$$

You can verify each of these by checking that

$$y^2 = x^3 + 2x + 3 \pmod{11}$$

If we take a point $P$ and write out $P, P+P, \ldots$, we get

  • $1P = (2, 2)$
  • $2P = (0, 5)$
  • $3P = (3, 5)$
  • $4P = (4, 3)$
  • $5P = (8, 6)$
  • $6P = (10, 0)$
  • $7P = (8, 5)$
  • $8P = (4, 8)$
  • $9P = (3, 6)$
  • $10P = (0, 6)$
  • $11P = (2, 9)$
  • $12P = O$

Compare this list of points with the list above. What do you notice? Every one of the points must be a point on the curve.

Next, how many points did we cycle through, which is the order?

This elliptic curve has order $\#E = |E| = 12$ since it contains $12$ points in its cyclic group.

There is a theorem called Hasse‘s Theorem: Given an elliptic curve module $p$, the number of points on the curve is denoted by $\#E$ and is bounded by $$p+1-2\sqrt{p} ≤ \#E ≤ p+1+2 \sqrt{p}$$

Interpretation: The number of points is close to the prime $p$.

  • $\begingroup$ Thanks, this makes sense, but if I choose $P=\left(4,3\right)$ then I get $\# E=3$ as $3P=\mathcal{O}$. What made you choose $P=\left(2,2\right)$? $\endgroup$
    – lioness99a
    Feb 13, 2017 at 15:13
  • $\begingroup$ I actually calculated all of them to see how many there were since the list of valid points was short. I chose one that had full period. Note there are algorithms to count the number of points on an EC: en.wikipedia.org/wiki/Counting_points_on_elliptic_curves. You might also want to read: math.stackexchange.com/questions/331329/… $\endgroup$
    – Moo
    Feb 13, 2017 at 15:53

I do not have enough reputation to comment, so I answer it here.

The comment says:

Thanks, this makes sense, but if I choose $P=(4,3)$ then I get $\#E=3$ as $3P=\mathcal{O}$, What made you choose P=(2,2)?

Because not all points are generators, however, only the generator point can generate all the points on the curve.

So, $P=(2,2)$ is one generator, but $P=(4,3)$ is not.

you can read this : https://www.entrust.com/wp-content/uploads/2014/03/WP_Entrust_Zero-to-ECC_March2014.pdf


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