# Why g is 2 and 3 to derive the lambda and beta values for endomorphism on the secp256k1 curve?

As you can seen here, in hex, N and P are:

N = FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFE BAAEDCE6 AF48A03B BFD25E8C D0364141

P = FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFE FFFFFC2F

The actual values of lambda and beta are easily verifiable and are:

β = 7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee

β2=851695D49A83F8EF919BB86153CBCB16630FB68AED0A766A3EC693D68E6AFA40

From Fermat's little theorem, if p is a prime number and g is a generator for the field Z/pZ, Z/nZ then:

(g ^ ((p - 1)/3)^3 = g ^ (p - 1) = 1
(g ^ ((N - 1)/3)^3 = g ^ (N - 1) = 1


β2 and λ2 can be generated by switching 2 and 3 in the equation, so we can generate 6 set of privet/public that group up in 3 rings.

Can some one explain why g choose to be 2 and 3?

what is the relation between 2 groups generated from λ which are (pvk, N-pvk) with each to be equal to N and 2N, and a group generated from β, which is (X coordinates) that sum of it be equal to P or 2P ?

Prime Curve (p), Prime Multiplier (N), Trace (P−N), Curve is Half. Multipliers (M1+M2)=N, y Coordinates + Inverse Y Coordinate = P.

6 Pubkeys are

 Pubkey = [x,y]  [x*beta%p, y]  [x*beta2%p, y] [x,p-y]  [x*beta%p, p-y]  [x*beta2%p, p-y]


6 Privatekeys are

pvk, pvk*lmda%N, pvk*lmda2%N, N-pvk, N-pvk*lmda%N, N-pvk*lmda2%N


is that possible to find the relation using mod ((N-1)+(P-1))/2?

There must be a third value such as λ and β to connect the abelian groups.