Title: Calculate ‘X’ coordinates that share the same ‘Y’
Post by: mcdouglasx on October 05, 2024, 05:20:00 PM
calculating the modular cubic root, determine all the values of X for a given Y coordinate. Usually, you can find 0, 1, 2, up to 3 values of X that share the same Y. update:import ecdsa
from ecdsa.ellipticcurve import Point
from sympy import symbols, Eq, solve
target = "0379be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798"
if target.startswith("04"):
Upub = target[2:]
y_value = int(Upub[64:], 16)
elif target.startswith("02") or target.startswith("03"):
x = int(target[2:], 16)
curve = ecdsa.SECP256k1.curve
y_square = (x**3 + curve.a() * x + curve.b()) % curve.p()
y_value = pow(y_square, (curve.p() + 1) // 4, curve.p())
if (target.startswith("03") and y_value % 2 == 0) or (target.startswith("02") and y_value % 2 != 0):
y_value = curve.p() - y_value
Upub = '04' + hex(x)[2:].zfill(64) + hex(y_value)[2:].zfill(64)
else:
raise ValueError("null")
p = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F
beta = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee
xcubed = (y_value * y_value - 7) % p
print("xcubed = 0x%x" % xcubed)
x1 = pow(xcubed, (p + 2) // 9, p)
x2 = (x1 * beta) % p
x3 = (x2 * beta) % p
UncompressedPublicKey1 = '04' + hex(x1)[2:].zfill(64) + hex(y_value)[2:].zfill(64)
compressedPublicKey1 = ('02' if y_value % 2 == 0 else '03') + hex(x1)[2:].zfill(64)
UncompressedPublicKey2 = '04' + hex(x2)[2:].zfill(64) + hex(y_value)[2:].zfill(64)
compressedPublicKey2 = ('02' if y_value % 2 == 0 else '03') + hex(x2)[2:].zfill(64)
UncompressedPublicKey3 = '04' + hex(x3)[2:].zfill(64) + hex(y_value)[2:].zfill(64)
compressedPublicKey3 = ('02' if y_value % 2 == 0 else '03') + hex(x3)[2:].zfill(64)
print("point 1:")
print(f"Uncompressed: {UncompressedPublicKey1}")
print(f"Compressed: {compressedPublicKey1}\n")
print("point 2:")
print(f"Uncompressed: {UncompressedPublicKey2}")
print(f"Compressed: {compressedPublicKey2}\n")
print("point 3:")
print(f"Uncompressed: {UncompressedPublicKey3}")
print(f"Compressed: {compressedPublicKey3}")
Example of 3 points that share Y: cPub: 0379be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
X= 79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
Y= b7c52588d95c3b9aa25b0403f1eef75702e84bb7597aabe663b82f6f04ef2777
cPub: 0379be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
H160: adde4c73c7b9cee17da6c7b3e2b2eea1a0dcbe67
Addr: 1GrLCmVQXoyJXaPJQdqssNqwxvha1eUo2E cPub: 03bcace2e99da01887ab0102b696902325872844067f15e98da7bba04400b88fcb
X= bcace2e99da01887ab0102b696902325872844067f15e98da7bba04400b88fcb
Y= b7c52588d95c3b9aa25b0403f1eef75702e84bb7597aabe663b82f6f04ef2777
cPub: 03bcace2e99da01887ab0102b696902325872844067f15e98da7bba04400b88fcb
H160: 516becf48ea6f02f20591504925d87bc62e9014f
Addr: 18RX2Lmi8cm2Bab8eABwyPgr9h8GSDjond cPub: 03c994b69768832bcbff5e9ab39ae8d1d3763bbf1e531bed98fe51de5ee84f50fb
X= c994b69768832bcbff5e9ab39ae8d1d3763bbf1e531bed98fe51de5ee84f50fb
Y= b7c52588d95c3b9aa25b0403f1eef75702e84bb7597aabe663b82f6f04ef2777
cPub: 03c994b69768832bcbff5e9ab39ae8d1d3763bbf1e531bed98fe51de5ee84f50fb
H160: 9226f4f38f59b60ad6d41fd78c7dcadb40cf1515
Addr: 1EKnKn1b4vQF81CxpBb2v2QiUhrjJsA4RK private_key = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364140
N = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
lambda_value = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72
private_key_x2 = (private_key * pow(lambda_value , 1, N)) % N
private_key_x3 = (private_key * pow(lambda_value , 2, N)) % N
print("Private key x1: 0x%x" % private_key)
print("Private key x2: 0x%x" % private_key_x2)
print("Private key x3: 0x%x" % private_key_x3)
Title: Re: Calculate ‘X’ coordinates that share the same ‘Y’
Post by: ecdsa123 on October 05, 2024, 06:06:09 PM
via sagemath
## Input
y = 0x3199555CE45C38B856C9F64AC6DB27000AB6CEA10CAD76B2B6E246C9A020E707
## Field parameters
# Field modulus
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
# Cube root of 1
beta = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee
## Actual code
xcubed = (y*y - 7) % p
print ("xcubed = 0x%x" % xcubed)
x = pow(xcubed, (p + 2) / 9, p)
print ("x1 = 0x%x" % x)
print ("x2 = 0x%x" % (x * beta % p))
print ("x3 = 0x%x" % (x * beta * beta % p))
Title: Re: Calculate ‘X’ coordinates that share the same ‘Y’
Post by: mcdouglasx on October 05, 2024, 08:00:32 PM
via sagemath
## Input
y = 0x3199555CE45C38B856C9F64AC6DB27000AB6CEA10CAD76B2B6E246C9A020E707
## Field parameters
# Field modulus
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
# Cube root of 1
beta = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee
## Actual code
xcubed = (y*y - 7) % p
print ("xcubed = 0x%x" % xcubed)
x = pow(xcubed, (p + 2) / 9, p)
print ("x1 = 0x%x" % x)
print ("x2 = 0x%x" % (x * beta % p))
print ("x3 = 0x%x" % (x * beta * beta % p))
thanks, maybe, if I find my old script I will post later to calculate their respective pk, when one of them is known.
Title: Re: Calculate ‘X’ coordinates that share the same ‘Y’
Post by: ecdsa123 on October 05, 2024, 08:37:33 PM
here u go:
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
E = EllipticCurve(GF(p), [0, 7])
G = E.point( (0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798,0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8)) # Base point
## Input
y = 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
## Field parameters
# Field modulus
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
# Cube root of 1
beta = 60197513588986302554485582024885075108884032450952339817679072026166228089408
## Actual code
xcubed = (y*y - 7) % p
print ("xcubed = 0x%x" % xcubed)
x = pow(xcubed, (p + 2) / 9, p)
print ("x1 = 0x%x" % x)
print ("x2 = 0x%x" % (x * beta % p))
print ("x3 = 0x%x" % (x * beta * beta % p))
x1=x * beta % p
x2=(x * beta * beta % p)
P=E.point((x,y))
P2=E.point((x1,y))
P3=E.point((x2,y))
assert G*78074008874160198520644763525212887401909906723592317393988542598630163514318 == P
assert G*37718080363155996902926221483475020450927657555482586988616620542887997980018 == P2
Title: Re: Calculate ‘X’ coordinates that share the same ‘Y’
Post by: mcdouglasx on October 05, 2024, 09:16:58 PM
here u go:
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
E = EllipticCurve(GF(p), [0, 7])
G = E.point( (0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798,0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8)) # Base point
## Input
y = 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
## Field parameters
# Field modulus
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
# Cube root of 1
beta = 60197513588986302554485582024885075108884032450952339817679072026166228089408
## Actual code
xcubed = (y*y - 7) % p
print ("xcubed = 0x%x" % xcubed)
x = pow(xcubed, (p + 2) / 9, p)
print ("x1 = 0x%x" % x)
print ("x2 = 0x%x" % (x * beta % p))
print ("x3 = 0x%x" % (x * beta * beta % p))
x1=x * beta % p
x2=(x * beta * beta % p)
P=E.point((x,y))
P2=E.point((x1,y))
P3=E.point((x2,y))
assert G*78074008874160198520644763525212887401909906723592317393988542598630163514318 == P
assert G*37718080363155996902926221483475020450927657555482586988616620542887997980018 == P2
I had already found my code but, thanks anyway.
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