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:
λ = 5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72
β = 7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee
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
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
There must be a third value such as λ and β to connect the abelian groups.