# Does secp256r1 or secp256k1 use a=-3?

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?

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.
• @WeCanBeFriends Note what $p$ is. Recall that $-3 \equiv p - 3 \pmod p$. – Squeamish Ossifrage May 1 '18 at 20:15