I saw a post here which I think states that for P256, a is equal to -3.

Why is elliptic curve parameter $a=-3$ somehow special

I am not sure if this is correct, or maybe I am misunderstanding something about elliptic curves?

I thought that a=0 for secp256r1 and some big number for secp256k1?

If we change the value to a=-3, then we no longer have secp256r1 and secp256k1 right?


1 Answer 1


The equation for secp256k1 is $y^2 = x^3 + 7$, so that $a = 0$ and $b = 7$.

The equation for secp256r1, also known as NIST P-256, is $y^2 = x^3 - 3x + b$, where $b$ is 41058363725152142129326129780047268409114441015993725554835256314039467401291; here $a \equiv -3 \equiv p - 3 \pmod p$ where $p = 2^{256} - 2^{224} + 2^{192} + 2^{96} - 1$ is the characteristic of the field underlying secp256r1.

If you change the curve parameters, you get a different curve.

  • $\begingroup$ I checked this link on pg15 by secg: secg.org/SEC2-Ver-1.0.pdf . It says that a= = FFFFFFFF 00000001 00000000 00000000 00000000 FFFFFFFF FFFFFFFF FFFFFF For secp256r1. Am I missing something? $\endgroup$ Commented May 1, 2018 at 20:00
  • $\begingroup$ @WeCanBeFriends Note what $p$ is. Recall that $-3 \equiv p - 3 \pmod p$. $\endgroup$ Commented May 1, 2018 at 20:15
  • $\begingroup$ Ohhh, so that number is the same as -3 in terms of mod. Thanks Squeamish! $\endgroup$ Commented May 1, 2018 at 20:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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