Deriving an ECDSA uncompressed public key from a compressed one

Question

I am currently trying to derive a Bitcoin uncompressed ECDSA public key from a compressed one.

According to this link on the Bitcoin wiki, it is possible to do so... But how?

To give you more details: as of now I have compressed keys (33-bytes-long) gathered on the bitcoin network.

They are of the following format: <1-byte-long prefix><32-bytes-long X>. From there, I would like to obtain an uncompressed key (65-bytes-long) whose format is: <1-byte-long prefix><32-bytes-long X><32-bytes-long Y>

According to this other link on the Bitcoin wiki, it should be as easy as solving the equation:

Y^2 = X^3 + 7

However, I cannot seem to get there. My value for Y is simply far-off. Here is my code (the value for the public key come from the Bitcoin wiki example):

import binascii
from decimal import *

expected_uncompressed_key_hex = '0450863AD64A87AE8A2FE83C1AF1A8403CB53F53E486D8511DAD8A04887E5B23522CD470243453A299FA9E77237716103ABC11A1DF38855ED6F2EE187E9C582BA6'
expected_y_hex = expected_uncompressed_key_hex[-64:]
expected_y_dec = int(expected_y_hex, 16)
x_hex = expected_uncompressed_key_hex[2:66]
if expected_y_dec % 2 == 0:
    prefix = "02"
else:
    prefix = "03"

artificial_compressed_key = prefix + x_hex

getcontext().prec = 500
test_dec = Decimal(int(x_hex, 16))
y_square_dec = test_dec**3 + 7
if prefix == "02":
    y_dec = - Decimal(y_square_dec).sqrt()
else:
    y_dec = Decimal(y_square_dec).sqrt()

computed_y_hex = hex(int(y_dec))
computed_uncompressed_key = "04" + x + computed_y_hex

For information, my outputs are:

computed_y_hex = '0X2D29684BD207BF6D809F7D0EB78E4FD61C3C6700E88AB100D1075EFA8F8FD893080F35E6C7AC2E2214F8F4D088342951'
expected_y_hex = '2CD470243453A299FA9E77237716103ABC11A1DF38855ED6F2EE187E9C582BA6'

Thank you for your help!

Answer 1

Here a sample code without any 3rd party python libs:

def pow_mod(x, y, z):
    "Calculate (x ** y) % z efficiently."
    number = 1
    while y:
        if y & 1:
            number = number * x % z
        y >>= 1
        x = x * x % z
    return number

# prime p = 2^256 - 2^32 - 2^9 - 2^8 - 2^7 - 2^6 - 2^4 - 1
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f

# bitcoin's compressed public key of private key 55255657523dd1c65a77d3cb53fcd050bf7fc2c11bb0bb6edabdbd41ea51f641
compressed_key = '0314fc03b8df87cd7b872996810db8458d61da8448e531569c8517b469a119d267'

y_parity = int(compressed_key[:2]) - 2
x = int(compressed_key[2:], 16)

a = (pow_mod(x, 3, p) + 7) % p
y = pow_mod(a, (p+1)//4, p)

if y % 2 != y_parity:
    y = -y % p

uncompressed_key = '04{:x}{:x}'.format(x, y)
print(uncompressed_key) 
# should get 0414fc03b8df87cd7b872996810db8458d61da8448e531569c8517b469a119d267be5645686309c6e6736dbd93940707cc9143d3cf29f1b877ff340e2cb2d259cf

refer to bitcoin talk: https://bitcointalk.org/index.php?topic=644919.0

Answer 2

You need to calculate in the field , which mostly means that you have to reduce your number to the remainder after dividing with p after each calculation. Calculating this is called taking the modulo and is written as % p in python.

Exponentiating in this field can be done more effectively than the naive way of just multiplying and reducing many times. This is called modular exponentiation. Python's built-in exponentation function pow(n,e,p) can take care of this.

The remaining problem is to find the square root. Luckily secp256k1 is chosen in a special way (), so that taking square roots is easy: A square root of x is .

So a simplified version of your code becomes:

import binascii

p_hex = 'FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F'
p = int(p_hex, 16)
compressed_key_hex = '0250863AD64A87AE8A2FE83C1AF1A8403CB53F53E486D8511DAD8A04887E5B2352'
x_hex = compressed_key_hex[2:66]
x = int(x_hex, 16)
prefix = compressed_key_hex[0:2]

y_square = (pow(x, 3, p)  + 7) % p
y_square_square_root = pow(y_square, (p+1)/4, p)
if (prefix == "02" and y_square_square_root & 1) or (prefix == "03" and not y_square_square_root & 1):
    y = (-y_square_square_root) % p
else:
    y = y_square_square_root

computed_y_hex = format(y, '064x')
computed_uncompressed_key = "04" + x_hex + computed_y_hex

print computed_uncompressed_key

Answer 3

The field of the elliptic curve is not over the field of real numbers. It's over a finite field modulo some prime.

For Secp256k1 the prime p = 2^256 - 2^32 - 2^9 - 2^8 - 2^7 - 2^6 - 2^4 - 1.

Thus: y^2= (x^3) + 7 (mod p)

There's no direct way to solve the equation, you would need to use Cipolla's algorithm: https://en.wikipedia.org/wiki/Cipolla%27s_algorithm