How to change the field of the elliptic curve to the new value N while keeping the default Secp256k1 parameters?

Question

I want to change the margin of the elliptic curve to the new N value, while doing so I want to keep the parameters Standard G-point Generator Secp256k1 that is, I want to replace: N = 115792089237316195423570985008687907852837564279074904382605163141518161494337

to the new value: N = 115792089237316195423570985008687907853269978629958289035461602417822422574827

leave the Secp256k1 parameters:

Gx = 55066263022277343669578718895168534326250603453777594175500187360389116729240
Gy = 32670510020758816978083085130507043184471273380659243275938904335757337482424

P = 115792089237316195423570985008687907853269984665640564039457584007908834671663

For the calculation, I use a Python script: but in the end I get the same result as in the standard Secp256k1 parameters

that is, with Private key 2, I get

Public Key:
04C6047F9441ED7D6D3045406E95C07CD85C778E4B8CEF3CA7ABAC09B95C709EE51AE168FEA63DC339A3C58419466CEAEEF7F632653266D0E1236431A950CFE52A
Public Key (compressed):
02C6047F9441ED7D6D3045406E95C07CD85C778E4B8CEF3CA7ABAC09B95C709EE5

the result is the same as in the field N = 115792089237316195423570985008687907852837564279074904382605163141518161494337

What exactly am I doing wrong because I have set a new field? N = 115792089237316195423570985008687907853269978629958289035461602417822422574827

What needs to be changed so that as a result the Public Key is different from the standard Secp256k1 parameters?

Code :

Pcurve = 2**256 - 2**32 - 2**9 - 2**8 - 2**7 - 2**6 - 2**4 -1 # The proven prime
Acurve = 0; Bcurve = 7 # These two defines the elliptic curve. y^2 = x^3 + Acurve * x + Bcurve
Gx = 55066263022277343669578718895168534326250603453777594175500187360389116729240
Gy = 32670510020758816978083085130507043184471273380659243275938904335757337482424
GPoint = (Gx,Gy) # Generator Point

N = 115792089237316195423570985008687907853269978629958289035461602417822422574827
privKey = 2

def modinv(a,b=Pcurve): #Extended Euclidean Algorithm/'division' in elliptic curves
    lm, hm = 1,0
    low, high = a%b,b
    while low > 1:
        ratio = high/low
        nm, new = hm-lm*ratio, high-low*ratio
        lm, low, hm, high = nm, new, lm, low
    return lm % b

def ECAdd(a,b): # Point addition, invented for EC.
    LambdaAdd = ((b[1] - a[1]) * modinv(b[0] - a[0],Pcurve)) % Pcurve
    x = (LambdaAdd * LambdaAdd - a[0] - b[0]) % Pcurve
    y = (LambdaAdd * (a[0] - x) - a[1]) % Pcurve
    return (x,y)

def ECDouble(a): # Point Doubling, also invented for EC.
    LamdaDouble = ((3 * a[0] * a[0] + Acurve) * modinv((2 * a[1]), Pcurve)) % Pcurve
    x = (LamdaDouble * LamdaDouble - 2 * a[0]) % Pcurve
    y = (LamdaDouble * (a[0] - x) - a[1]) % Pcurve
    return (x,y)

def ECMultiply(GenPoint,privKeyHex): #Double & add. Not true multiplication
    if privKeyHex == 0 or privKeyHex >= N: raise Exception("Invalid Private Key")
    privKeyBin = str(bin(privKeyHex))[2:]
    Q=GenPoint
    for i in range (1, len(privKeyBin)):
        Q=ECDouble(Q);
        if privKeyBin[i] == "1":
            Q=ECAdd(Q,GenPoint);
    return (Q)

publicKey = ECMultiply(GPoint,privKey)
print "Private Key:";
print privKey; print
print "Public Key (uncompressed):";
print publicKey; print
print "Public Key (compressed):";
if publicKey[1] % 2 == 1: # If the Y coordinate of the Public Key is odd.
    print "03"+str(hex(publicKey[0])[2:-1]).zfill(64)
else: # If the Y coordinate is even.
    print "02"+str(hex(publicKey[0])[2:-1]).zfill(64)

Answer 1

I want to change the margin of the elliptic curve to the new N value, while doing so I want to keep the parameters Standard G-point Generator Secp256k1

What do you mean by "margin"?

What you're asking for is a contradiction. The generator order N is a property of the curve and its generator. It is not something that can be chosen independently.

It is the answer to the question "After how many times adding the generator to itself do I end up with the point at infinity". It is completely determined by the curve equation (Y^2=X^3+7 for secp256k1) and the generator (which must be on that curve). N is computed using Schoof's algorithm.

If you want N to be different, you will need to pick a different generator. In the case of secp256k1, where the curve order is prime, every point has the same order, so another choice won't help (except the point at infinity, which by definition has order 1).

So, if you want a system where the generator order is the N value you desire, you'll need a different curve as well. Finding a curve which admits a generator with a specific order is a very nontrivial problem. I don't know of any generic way of doing so.

to the new value: N = 115792089237316195423570985008687907853269978629958289035461602417822422574827

Where does that number come from? Does it have some special properties that make you believe a curve with that order should exist?

For the calculation, I use a Python script: but in the end I get the same result as in the standard Secp256k1 parameters

This is completely expected, as you're not using N in the computation (nor can you). You're just computing generator+generator.

Answer 2

The elliptic curve is secp256k1: y² = x³ + 7 (mod p)

where the field size is

p = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F.

The curve order of secp256k1 is:

n = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141

The public key P (point) is calculated by multiplying the private key d (scalar) by the generator point G (point) where

P = d (mod n).G

The n (N) you state is not the field size p, it is the curve order

n = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141

This curve order n is such that nG = point at infinity. This is a property of the curve and cannot be changed.

Thanks to Pieter Wuille for the corrections.