I hear that there should be a point at infinity on secp256k1. I wounder how to calculate it and what does it even mean. I tried to calculate it as $P_{inf}=P+(-P)$ but this gives different results for different P. Are there more than one point at infinity?

Geometrically thinking on an elliptic curve, adding two points that are symmetric against x-axis, should not be feasible because any vertical line on x-y plane will only intersect the curve at TWO points not THREE.


3 Answers 3


Point at Infinity

Yes, there is a point at infinity $O$ (or denoted as $\infty$) on Elliptic Curves (EC) as the identity element of the Elliptic curve group. Let the EC be given in Weierstrass equation;

$$y^2 = x^3 +a x + b$$

$O$, the point at infinity can have different values according to the underlying coordinate system;

Coordinate Systems

For an elliptic curve over $\mathbb{F}_p$;

  • In Homogeneous Coordinates (also known as Projective Coordinate), the points are defined as $(x:y:z),z\neq 0$ corresponding to affine point $(\frac{x}{z},\frac{y}{z})$. The point at infinity is $(0:1:0)$ and negative value of $(x:y:z)$ is $(x:-y:z)$
  • In Jacobian Coordinate the points are defined as $(x:y:z),z\neq 0$ corresponding to affine point $(\frac{x}{z^2},\frac{y}{z^3})$. The point at infinity is $(1:1:0)$ and negative value of $(x:y:z)$ is $(x:-y:z)$

Now, what is $O$ in affine coordinates? Try to convert $O$ back into an affine coordinate! Well, we cannot represent it! See this answer Thus, we simply say it is $O$. But, magically, it can be encoded in implementations.

The Group Law

The points on an elliptic curve form an additive group with an identity $O$. For geometrical meaning please see the below image from Wikipedia Elliptic Curves The group law.

Geometric interpretation of addition on a Weierstrass curve

The Group Law on Affine Coordinates

Arithmetically can define the addition rules in affine coordinates. These are derived from the line intersection and tangent equations. The formulas are;

Let $P=(x_1,y_1)$ and $Q=(x_2,y_2)$ be two point in the elliptic curve.

  1. $P+O=O+P=P$
  2. If $x_1 = x_2 $ and $y_1 = - y_2$, that is $Q =(x_2,y_2)=(x_1,−y_1)=−P$, then $P+Q = P + (-P) = O$
  3. If $Q \neq -P$ then the addition $P+Q = (x_3,y_3)$ and the coordinate can be calculated by;

\begin{align} x_3 = & \lambda^2 -x_1 - x_2 \mod p\\ y_3 = & \lambda(x_1-x_3) -y_1 \mod p \end{align}

$$ \lambda = \begin{cases} \frac{y_2-y_1}{x_2-x_1}, & \text{if $P \neq Q$} \\[2ex] \frac{3 x_1^2+a}{2y_1}, & \text{if $P = Q$} \\[2ex] \end{cases}$$

The Group Law on projective Coordinates

Let $P_i = (x_i : y_i : z_i ), i = 1, 2$, be points on the elliptic curve $$y^2 z = x^3 + Axz^2 + Bz^3.$$ Then $$(x_1 : y_1 : z_1 ) + (x_2 : y_2 : z_2 ) = (x_3 : y_3 : z_3 ).$$ The formulas are;

  1. $P_1 \neq \pm P_2$
  • $u = y_2 z_1 − y_1 z_2,$
  • $v = x_2 z_1 − x_1 z_2,$
  • $w = u^2 z_1 z_2 − v^3 − 2v^2 x_1 z_2,$ then

$$x_3 = vw, \quad y_3 = u(v^2 x_1 z_2 − w) − v^3 y_1z_2, \quad z_3 = v^3 z_1 z_2$$

  1. $P_1 = P_2$
  • $t = a z_1^2 + 3x_1^2,$
  • $u = y_1 z_1,$
  • $v = u x_1 y_1,$
  • $w = t^2 − 8 v,$

$$x_3 = 2uw, \quad y_3 = t(4v − w) − 8y_1^2 u^2 , \quad z_3 = 8 u^3.$$

For Jacobian Coordinate operations see this link.

  • 1
    $\begingroup$ In 2 you mean P + Q = O (or P + -P) $\endgroup$ Jan 6, 2019 at 10:13
  • $\begingroup$ @kelalaka the Wikipedia link you have supplied is saying that if P and Q are opposites of each other, we define $P + Q = O$. Why we cannot conclude $P+(-P)=O$ is always true? By the way what is the random looking output value is that my computer program in python is giving as output? finally is it possible to calculate $O$ so that for any randomly given $P$ the computer program can calculate $P+O=P$? $\endgroup$
    – PouJa
    Jan 6, 2019 at 10:57
  • $\begingroup$ The points are forming an additive group with identity $O$ so $P+(-P) = O$. Wikipedia also says the opposite of a point $P$ is $-P$. For the programming part, unfortunately, off-topic here. Please ask a question on StackOverflow. $\endgroup$
    – kelalaka
    Jan 6, 2019 at 11:17

What does (the point at infinity) even mean?

That's an abstract point/concept required so that the addition of points on the Elliptic Curve is a group law, including

  • Addition of any two elements $P$ and $Q$ of the group is an element of the group
  • There is a neutral element $\mathcal O$ such that for all $P$, $P+\mathcal O=P=\mathcal O+P$.

A possible definition: the point at infinity is a single extra element $\mathcal O$ added to the curve and the rules of point addition, such that for all $P$ on the curve (including the point at infinity $\mathcal O$), it holds $P+\mathcal O=P=\mathcal O+P$.

We can then define $-P$ as the point $Q$ such that $P+Q=\mathcal O=Q+P$, and for all $P$ it holds $P+(-P)=\mathcal O=(-P)+P$.

How to calculate (the point at infinity)

One does not, much like one does not calculate $+\infty$.

One may define a conventional representation of the point at infinity in the Cartesian coordinate system by picking any coordinate not on the curve, e.g. $(x,y)=(0,0)$ assuming $b\ne0$. With this convention, a point $(x,y)$ is on the curve if $(x,y)=(0,0)$ or if $y^2 = x^3 +a x + b$. And the addition law is extended as

$$(x_0,y_0)+(x_1,y_1)=\begin{cases} (0,0)&\text{ if }(x_0,y_0)=(x_1,-y_1)\\ (x_0,y_0)&\text{ if }(x_1,y_1)=(0,0)\\ (x_1,y_1)&\text{ if }(x_0,y_0)=(0,0)\\ \text{per usual rules }&\text{ otherwise} \end{cases}$$

As noted in this comment, the point at infinity has a more natural representation in projective coordinates.

A correct implementation of point addition should give a representation of the point at infinity $\mathcal O$ when asked to compute $P+(-P)$ for $P$ any point on the curve. Note that in Cartesian coordinates, if $P=(x,y)$ and $P\ne\mathcal O$, then $-P=(x,-y)$ or $-P=(x,(p-y)\bmod p)$ depending on conventions. The representation of $\mathcal O$ is not necessarily unique, therefore $P+(-P)=Q+(-Q)$ does not necessarily hold.

  • 1
    $\begingroup$ I like that angle of connecting with the group law that requires neutral element of the group operation. Another nice connection would be through projective coordinates - for projective coordinates (x,y,z) we have an affine representation (x/z. y/z) and the only point that cannot be represented that way is the case when z=0 which is exactly the point at infinity. $\endgroup$
    – Kris
    Feb 18, 2020 at 13:10

For the secp256k1 curve, there is a point of infinity which is N the total points for this curve. To calculate use scalar multiplication of N and the base generator point for the curve. Another way is to go 2coin.org, go to the private key hex tab, enter the number 1 for the private key press enter. Then subtract by pressing -1 you will be at infinity, press one more time, you are at N-1.

  • $\begingroup$ That result does not satisfy the curve equation. It means that it is not on the curve! $\endgroup$
    – PouJa
    Feb 24, 2020 at 16:06

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