For Bitcoin's ECDSA curve (secp256k1, where $a=0$, $b=7$), why can't the generator point's first coordinate be $x=0$? That is, the point on the curve would be $(0,y)$ where $y$ satisfies $y^2 = 0^3 + 7=7$.

Are there any elements other than $0$ which can't be used as a generator point's $x$ coordinate?

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    $\begingroup$ That point doesn't exist. For the prime modulus used by secp256k1, 7 is a quadratic non-residue, meaning y^2 = 7 has no solution. $\endgroup$
    – user13741
    Jan 19, 2015 at 6:09
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    $\begingroup$ For an elliptic curve, about half the x values have two associated points, the other half of the x values has none (they correspond to the twist of the curve). $\endgroup$ Jan 19, 2015 at 9:31
  • $\begingroup$ @user13741 The point certaily exists, but it is not on the curve. ;) $\endgroup$
    – fkraiem
    Dec 17, 2015 at 7:52

2 Answers 2


Here is a plot of secp256k1 over the reals. That I'll use for illustration purposes.

enter image description here

What happens at $x=-3$, well, we get $y^2 = (-3)^3+7$ which reduces to $y^2=-20$. Solving for $y$ we get $y=\sqrt{-20}$, which is not in the reals, so, we cannot have a point on that curve for $x=-3$.

This is what is happening. In the case of the real secp256k1 used in bitcoin, we are using a finite field instead of the reals. In the reals, the equation $y^2=x^3+7$ is not defined for the entire domain. Similarly, over the finite field, the equation is not defined over the entire field.

In other words, in order for there to be a point (actually 2 points) on the curve at $x$, we must be able to take the square root of $x^3+7$ in the field. As to the number of valid $x$ values, see this question on Math.SE.


The point simply is not on the curve!


The curve secp256k1 has no point such that the x-coordinate is 0. There is no integer which its square is 7 modulo FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFE FFFFFC2F.

I hope this helps to understand why there is no such a point.

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    $\begingroup$ Need more explanation in your answer for me to give it a +1. $\endgroup$
    – mikeazo
    Jan 20, 2015 at 15:49
  • $\begingroup$ @mikeazo My answer is updated. $\endgroup$
    – Nayef
    Feb 4, 2015 at 8:05

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